Si ATE BOARD 

EXAMINATIONS 



QUESTIONS AND 
ANSWERS 




Class ^ " E ^^f 

Book r^ ^ 

CDPXRIGHT DEPOSIT. 



State Board 
Examinations 

Questions and Answers 

By CV HENRY BROWN, M. D. 

Formerly Physician Philadelphia Hospital; 
Professor Principles and Practice of Op- 
tometry; Author the Optician' s Manual, 
Vols. I and II; Clinics in Optometry, Etc. 




19 19 



The keystone publishing Co. 

PHILADELPHIA, U. S. A, 






COPYRIGHT, I919, BY 
THE KEYSTONE PUBLISHING CO. 



OCT jO Idl9 






-1. . 


©CU536373 


^/^)' 



Preface 



The one thousand questions answered in this volume were 
carefully selected from the examination papers set by the State 
Boards of Examiners in Optometry. The book was compiled to 
meet the needs of those who may find it compulsory or advisable 
to take a State Board examination, and for optometry students 
generally. 

The questions are probably more fully answered than a 
State examining board would expect, the idea being to give com- 
plete information on the special subject brought out by each 
question. 

The contents are classified under different headings, which 
gives the work the character of a text-book and facilitates study 
and reference. As the questions were selected with a view to 
avoiding repetition, the book will be found to cover thoroughly 
every paper set by any of the State Examining Boards. 



Contents 



Theoretic Optics 9 

Practical Optics ' 97 

Theoretic Optometry 137 

Practical Optometry 174 

Pathological Optometry 218 

Physiological Optics 236 

Anatomy of the Eye , 272 

Ophthalmoscopy 283 

Retinoscopy 293 

Physiology of Vision 321 

Pathological Conditions ! 346 



Theoretic Optics 

A coin one inch in diameter is held twelve inches from a lens 
which has a principal focus of four inches. Where will the image 
he formed and what will he its size? Show the calculation. 

Parallel rays of light passing through this lens would be 
brought to focus at a distance of four inches. If the coin was 
placed at twelve inches the rays would be divergent and would 
reduce the power of the lens and throw the focus farther away. 
As these inch numbers must be expressed in fractions, the problem 
would be worked out by the usual formula, which in this case 
would be yi —1/12 = 1/6. So that the point where the image 
would be six inches from the lens and on the opposite side from 
the object. 

Since the object is at a distance of twelve inches from the 
lens and the image at a distance of six inches from it, which is 
just one-half the distance, the size would be one-half the size 
of the object, or one-half inch. 



Transpose into their plus and plus equivalents: 

a + 1.50 D. sph. C - .37 D. cyl. X 10° 
b + 5.00 D. sph. C - 4.50 D. cyl. X 150° 
c + 4.25 D. sph. C - 1.75 D. cyl. X 35° 

The rules that guide us are as follows: 

The sphere is obtained by the algebraic addition of the sphere 
and cylinder. 

The cylinder remains the same, but its sign and axis is 
changed. 

Using these rules the results will be 

a + 1.13 D. sph. C -f .37 D. cyl. axis 100° 
h + .50 D. sph. C + 4.50 D. cyl. axis 60° 
c + 2.50 D. sph. C + 1.75 D. cyl. axis 125° 



10 



State Board Examinations 



Transpose into sph. cyl. form: 

a + 2.25 D. cyl. X 10° O -\- 2.25 D. cyl. X 100° 
b - 1.75 D. cyl. X 70° C + 3.00 D. cyl. X 160° 
c + .75 D. cyl. X 55° C^ - 1.75 D. cyl. X 145° 

a. As this shows the same power in both meridians, and as 
these meridians are at right angles, the transposition would 
result not in a sphero-cylinder, but in a simple sphere of the same 
power. The answer is + 2.25 D. sphere. 

h. The rules for the transposition of cross-cylinders are as 
follows : 

1. Take either one of the cylinders for the sphere. 

2. Take the algebraic difference for the cylinder, retaining 
the sign and axis of the cylinder that was not taken for the 
sphere. 

Illustrate hy three simple diagrams the law of reflection for a 
luminous point centrally placed before a plane, a convex and a 
concave mirror respectively. 




Fig. 1 



The law of reflection is that the angle of reflection is the same 
as the angle of incidence. This law is illustrated in the preceding 



Theoretic Optics 11 

diagram of a plane mirror where A B \?> the incident ray proceed- 
ing from the candle and B C the reflected ray. 

The law of reflection is again illustrated in the second 
diagram, Fig. 2, representing a concave mirror where the parallel 
rays proceeding from A and B and striking C and D are reflected 
according to the law and meet at F. 




Fig. 2 

The law of reflection is also illustrated in the third diagram, 
Fig. 3, representing a convex mirror, where the parallel rays 
proceeding from A and B and striking C and D are reflected 
according to the law and diverge as if coming from F. 



Give the formula that defines the conjugate focal distances u 
and V for a spheric mirror, expressed in terms of the object distance u, 
the image distance v and the radius r. 

The refractive powder of a lens is inversely proportional to 
its focal length. For example if F represents the focal length 
1/F would represent the refractive power of a lens. 

\i u represents the distance of the object 

If V represents the distance of the image 

If/ represents the distance of the principal focus 

If r represents the radius of curvature, 
then we have the formula^ 

1,1 1 2 

= — or — 

u V f r 

If index is approximately 1.50, the focal length will be twice 



12 



State Board Examinations 



the radius of curvature (2 r) for piano lenses, while in biconvex 
lenses the focal length is equal to the radius (F = r). 



F<:- 




FiG. 3 



When all of the distances, u of the object, v of the image and 
the radius of curvature r from the pole on the axis of a spheric 
mirror are counted positive in the direction opposite to light incidence, 
is the mirror convex or concave and is the image real or virtual? 

In this diagram, Fig 4, the rays proceed from B, which is 
situated beyond the center of curvature C, and are focussed at D. 




Fig. 4 

These conditions correspond to those mentioned in the 
question and the mirror is concave and the image is real. 

Give the value of the principal focal distance f of a spheric 
mirror, as deduced from the general formula for conjugate focal 
distances. 

The formula is J: _i_ Ji L 

^ + ^ ~ f 

If the distance of the object is 30 inches and the distance 
of the image is 10 inches, then the principal focus is 7^, which 
we get by substituting values as follows: 

30 ^ 10 7W 



Theoretic Optics 



13 



Construct a diagram in which the linear dimensions of the 
object and the image i are represented by lines perpendicular to 
the axis at a greater distance than the focus of a concave mirror and 
deduct therefrom the ratio of magnification expressed in terms of 
the focus f and the distance u of the object. 

In this diagram, Fig. 5, ^ F £ is a concave mirror, F X is 
the principal axis which passes through the center of curvature 
C and the principal focus F. B D is a. luminous object in front 
of the mirror at a distance D V from the mirror. 




Fig. 5 



One of the rays of light from the point B will pass through 
the center of curvature C and will strike the mirror at the point 
E. As it is incident in the same direction as a radius of curvature 
it is perpendicular to the arc of the circle and hence will be 
reflected back along the same line as it came. 

Another ray of light from the point B, running parallel to 
the axis strikes the mirror at A and is reflected at such an angle 
as to pass through the principal focus F. 

In its further progress it intersects the first-mentioned ray 
at B\ which is, therefore, the image point of B. Consequently, 
D' B' is a real and inverted image oi B D and it is apparent from 
the diagram that the image is smaller than the object. 

The reverse of the above would also be true. Suppose 
D' B' was the object a parallel ray B' H after reflection would 
pass through F; and another ray from B' drawn through the 
center of curvature C would be reflected back upon itself and 
the two rays would intersect at B. Consequently B D would 
be the magnified image of D' B\ 



14 



State Board Examinations 



If i represents the image and o represents the object, the 
magnification or minification would be expressed by the fraction 
i/o. 

The triangles B D F and VHP are similar and their cor- 
responding sides are proportional. 

D' V is distance of object from mirror = u. -F F is distance 
of focus from mirror = /. 

D' B' = object = 

B D = image = i 
therefore 

i B D 



D' B' 



Also 



i D F, u - f 



What is meant by the index of refraction? 

The index of refraction is the numerical expression which 
indicates the refractivity of a medium as compared with air, 




Fig. 6 

which is taken as the standard or unit or 1, and hence every other 
medium being denser has an index greater than 1. The index 
of refraction depends upon the relation or ratio which constantly 
exists between the sine of the angle of incidence and the sine of 
the angle of refraction. 



Theoretic Optics 15 

In order to illustrate the course of a ray of light as it passes 
from one medium to another of different density, Fig. 6, let G H 
represent a refracting surface separating air with an index of 1.00 
from glass with an index of 1.50. Let / i^ be a ray incident on 
the surface of the point K, to which the line L K M is the normal 
or perpendicular ; then J K Lis the angle of incidence and P K M 
is the angle of refraction. 

If we represent the angle of incidence by the letter i and the 
angle of refraction by the letter r, then the formula for ascertain- 
ing the index of refraction of any substance is expressed as 
follows : 

sin i _ L J 

sin r P M 

U L J equals 3 and P M equals 2, then the first divided by 
the second equals 1.50 representing the index of refraction of 
the glass as compared with the standard taken or air. 



Give the general formula that defines the conjugate focal distance 
u and V and for a convex refracting surface whose radius is r and 
whose refractive index is n, the object being at a distance u and the 
image at a distance v from the pole; also state in which direction 
from the pole the distances on the principal axis are counted positive 
or negative with respect to the direction of incidence. 



Fig. 7 

In this figure, Fig. 7, the rays proceeding from the object 
are converged after refraction and form the image /. Each is a 
real focus and the two are called conjugate foci. The convexity 
of the surface is turned toward the incident wave. 

r = radius. 

n = refractive index rarer medium. 

n' = refractive index of denser medium. 



16 State Board Examinations 

u = distance of object. 
V = distance of image. 



n n 



AO ' A I r 

Substituting u for A and V for A I respectively we have 



n n _ n — n 
u V r 



This equation expresses the relation between the conjugate 
focal distance, A and A I. U is positive when measured in a 
direction opposite to the incident light and negative when meas- 
ured in the other direction. 




Fig. 8. (See opposite) 

In the above formula which value must be given to u so as to 
obtain the second principal focal distance v, commonly designated 
as fi? 

If is situated at such a distance that its rays are parallel, 
the distance A or u must be regarded as infinity and then n/u 
must equal zero. 

Making this substitution in the foregoing equation we have 



V r 

from which we derive the corresponding value of v or /2. 



What is the direction of the refracted ray with respect to the 
second principal focus when the incident ray is parallel to the axis 
of a spheric refracting surface? 



Theoretic Optics 17 

From the formula given in the previous answer we ascertain 
the focusing point for rays that are parallel before refraction. 
The distance of this point from the surface is the posterior or 
second principal focal distance and is often denoted by the letter 
/2, the value of which is derived from the equation just mentioned. 



Show in a simple diagram, by means of the principal lines of 
direction, the positions of the first and second principal foci, Fi 
and Fi, with respect to a luminous point slightly above the principal 
axis, nearer than the focus and located to the right and in front of 
a concave refracting surface. 

Let A represent a luminous point, from which a parallel 
ray of light after refraction by the concave surface will diverge 
as if proceeding from the point f^ which is the second principal 
focal distance. A divergent ray from A striking, the concave 
refracting surface at B will after refraction be parallel to the axis 
and appear to proceed from the point F^. (Figure 8, opposite.) 



From the general formula for conjtigate focal distances, 

V u^ \ ri Ti / 

for a thin lens having a radius of curvature ri for the first surface 
{exposed to incidence) and a radius r2 for the second surface, give 
the value of the second principal focal distance f^. 

If tc equals infinity, then 

— or7^= {n — 1)\ I 

1^ f- \ri r-i) 



It being true that the first and second principal focal distances 
/i and /2 for a thin lens are equal, give the equation from which the 
focal length f may be deducted. 

If/ be the focal length of a thin convex lens, then 



f^^"-)^-^) 



18 State Board Examinations 

Or if each surface has the same radius of curvature the formula 
may be written 



What are the laws of refraction? 

A ray of light striking a medium perpendicularly is not bent. 

A ray of light striking a medium obliquely in passing from 
a rare to a denser medium is bent toward the perpendicular. 

A ray of light striking a medium obliquely in passing from 
a dense to a rarer medium is bent from the perpendicular. 



Define principal focus, ordinary focus and conjugate foci. 

The principal focus is the focal point for parallel rays. 

An ordinary focus is the meeting point for rays which di- 
verge from some point inside of infinity and are brought together 
at some point beyond the principal focus. 

Conjugate foci are interchangeable, so that the rays which 
diverge from one will always converge to the other. 



What two factors determine the dioptric power of a lens? 
Index of refraction and radius of curvature. 



When light passes from a medium which is optically dense to 
one that is optically rare, what change in the direction of the light 
is there? 

If the ray of light strikes the medium perpendicularly, it 
passes unchanged; if it strikes it obliquely it is bent away from 
the perpendicular. 

What is the rule for figuring the radius of curvature of lenses, 
the index of refraction being given, and the focal power of the lens? 

The radius of curvature divided by the index of refraction 
less unity, equals the focus of a plano-convex lens. Hence the 
index of refraction, less unity, multiplied by the focal power of 
the lens, equals the radius of curvature. 



Theoretic Optics 19 

Suppose the index of refraction is 1.50 and the focal length 
of the lens 20 cm., then we have 

(1.50 — 1) X 20 = 10 cm., which is the radius of curvature. 



The focal length of a convex lens is 12 cm.; an object is placed 
20 cm. in front of the lens. Where will the image be and will it be 
erect or inverted? Show the process of the figuring. 

If the focal length of the convex lens is 12 cm., the refractive 
power of the lens is 8.33 D. The focal length of curve refers to 
parallel rays before entering the lens. If such rays are divergent 
the focus is thrown farther away, according to the degree of 
divergence. 

If the rays proceed from a distance of 20 cm. they would 
have a divergence of 5 D., which must be subtracted from 8.33 D., 
leaving 3.33 D., which means a focal distance of 12 inches or 
30 cm. 

Or using the reciprocals we have the following equation: 

12 20 ~30 



Having a biconcave lens, it is found that using one surface 
as a mirror the image of a distant object is at 8 inches from the 
surface; while testing the other surface in the same way, the focus 
is 4 inches. What will be the power of the lens if the refractive 
index of the glass is 1.50? 

Inasmuch as the principal focus of a mirror is one-half the 
radius, therefore an 8-inch focus would indicate a 16-inch radius, 
or a 2.50 D. value; and a 4-inch focus an 8-inch radius or 5 D. 
value. 

Now the formula for focal distance is as follows: 



F 
or substituting values: 



' ^^-H-^^^) 



' = a.5o-i)(-l+-J-^) 



F 



20 State Board Examinations 



or 



or 



^ = {.50) (- 5D. + - 2.50 D.) 



i = - ^-7^ D. 



At what distance will parallel rays he focused by a concave 
mirror whose radius of curvature is 17 inches? 

The principal focus of a concave mirror is always equal to 
half the radius of curvature; therefore, in this case, parallel rays 
will be focused at 8>^ inches. 



What is the radius of curvature of a plus 1.50 lens, crown glass? 

The index of refraction is not mentioned; it may vary from 
1.48 to 1.56 or even 1.60, but for sake of illustration we will 
say it is 1.50. 

Then we have focus multiplied by index less unity equals 
radius of curvature, or substituting figures 
26 X (1.50 - 1) = 13 

Therefore, 13 inches is the radius of curvature if the lens is 
piano, as proved by the rule that the focal length of a plano- 
convex lens is equal to twice the length of its radius. 

If the lens is biconvex we have 

26 X 2 (1.50 - 1) = 26 

Here 26 inches is the radius of curvature as proving the rule 
that the focal length of a biconvex lens is equal to its radius. 



What radius of curvature must he ground on the two surfaces 
of a periscopic lens, one of the surfaces to he on a radius of 5 cm., 
the total dioptric power to he 8 D., and the refractive index to he 
1.50? How is it worked out? In inches or millimeters? 

If the radius of curvature of one surface is 5 cm. then accord- 
ing to the rule that the focal length is twice the radius, the focal 
length of this surface is 10 cm. which means that the value of 
this surface is + 10 D. 



Theoretic Optics 21 

Now then, if the total power of the lens is + 8 D., the 
opposite must show a — 2 D. curve, or 20 inches. If according 
to the rule this is twice the length of the radius, the latter must 
be 10 inches. 

If the index had been other than 1.50 the results would have 
been slightly different. 

// a point of light is at a distance of 16 inches from a + 4.50 
D. lens, where will the light he focus sed? 

The rays diverging from a point of light at a distance of 
16 inches will require + 2.50 power to overcome this divergence 
and make them parallel. This leaves + 2 power of the + 4.50 D. 
lens to act on these parallel rays, which will be brought to a 
focus at 20 inches. 

A 2 D. lens decentered 5 mm. will produce how much prismatic 
action? 

According to the standard there is 1° prismatic power 
developed for every 1 D. of refractive power with a decentration 
of 10 mm., and >^° for every 1 D. lens with a decentration of 
5 mm. 

The question asks how much prismatic action. We do not 
know whether the framer of the question had in mind prismatic 
power or deviation, but to cover both phases the answer would 
be as follows: 1° of prismatic power and as the angle of deviation 
is one-half the principal angle, >^° of deviation. 



Object is 10 inches high and its distance from the mirror is 
40 inches; the image formed by the mirror is 5 inches distant; what 
is the height of the image? 

The height of the image bears the same relation to the 
height of the object as the distance of the image bears to the 
distance of the object. 

Let X represent height of image which it is desired to find 
and using the known values we have 

X : 10 :: 5 : 40 
10 X 5 50 ,^ . 
orx= ^o-=40=^^ 



22 State Board Examinations 

Object is 80 inches from a mirror; its image is 20 inches from 
mirror and is 2 inches high. How high is the object? 

The same proportion holds good as in previous question. 
Using X for the unknown quantity and substituting the values 
as given we have 

20 : 80 :: 2 :x 
Then 

^ 80 X 2 ^ 160 ^ 8 
^ ~ 20 ~ 20 ~ 1 

Or :r = 8 inches, which is the height of the object. 



The object is 24 inches wide and its image is 3 inches wide. 
The distance of the object from the mirror is 32 inches; what is the 
distance of the image? 

The width of the object bears the same relation to the width 
of the image as the distance of the object bears the distance of 
the image. 

The first three values are given and the fourth or the un- 
known will be represented by x, and then we have the following 
proportion: 

24 : 3 :: 32 : X 

Then 

^ 32 X 3 ^ 96 ^ 4 
^ 24 24 1 

Or X = 4 inches, which will be the distance of the image. 



Let the image be 2 inches high and the object 20 inches high; 
let the distance of the image be 4 inches; what will be the distance 
of the object from the mirror? 

The size of the image is to the distance of the image as the 
size of the object is to the distance of the object, which may be 
expressed as follows: 

2 in. : 4 in. :: 20 : x 

Then 

4 X 20 80 



40 inches 



which is the distance of the object. 



Theoretic Optics 23 

hi the preceding example what will be the power of the mirror 
if it is convex? 

The general formula for mirrors Is as follows: 

u ^ V F 

It must be borne in mind that the formula for a convex 
mirror is the same as that for a concave mirror, but in working 
with these formulae it is necessary to remember that in the 
case of the convex mirror v is always a negative quantity, because 
the focus of a convex mirror is not real but virtual. 

Hence the formula would be 

u^ V F 

or substituting the values given in the previous question, 

1 D. + - 10 D. = - 9 D. 
which is the negative focus or power of the convex mirror. 



Solve the same problem by the formula, but the mirror to be 
concave. 

Here the quantities are all positive, and using the values 
given in the previous question expressed in diopters, we have 

1 D. + 10 D. = + 11 D. 
as the positive focus or power of the concave mirror. 



A ray of light is incident on a plane surface at an angle of 60° 
with the normal; the refracted ray forms an angle with the normal 
of 45°. The sine of 45° is 0.707, and the sine of the angle 60° is 
0.866; how is the index of refraction found and what is it in this 
particular case? 

To find the index of refraction of any substance as compared 
with air, it is necessary to divide the sine of the angle of incidence 
by the sine of the angle of refraction. In this case we divide the 
sine of the angle of incidence (0.866) by the sine of the angle of 
refraction (0.707) and the result of 1.225 approximately as the 
index of refraction in this case. 



24 State Board Examinations 

What is the relation of size of image to object as compared 
with the distance of these from a lens, and is there any difference in 
the rule with a convex and a concave lens? 

The size of the image is to the size of the object, as the dis- 
tance of the image is to the distance of the object. This appUes 
to both convex and concave lenses, and when any three of these 
values are known the fourth can be figured out. 



What effect has obliquity on the image of a point of light, such 
as an illumined pinhole in a sheet of black paper when formed by 
a thin convex lens? 

The production of a more or less irregular blur in place of 
a true point image. There is an increase in the refractive power 
of all meridians of the lens, but especially so in the meridian at 
right angles to the axis of rotation, thus developing the effect 
of a stronger sphere with the addition of a cylindrical element. 



The focus of a concave mirror is 20 inches; where will be the 
image of an object placed at the following distances from the mirror: 
20 inches, 40 inches, infinity? 

If an object be placed at 20 inches from a 20-inch concave 
mirror; in other words if the object be placed at the focal distance 
of a concave mirror its image will be at infinity. 

If the object be placed at 40 inches the mirror must first 
overcome the divergence of rays proceeding from this distance, 
leaving a sufficient convergence to bring the rays to a focus at 
40 inches. 

If the object was placed at infinity the image would be 
formed at the focal distance of the lens, which is 20 inches. 



Where must an object be placed to obtain a real image four 
times the size of the object, using a + 2.50 D. lens? 

If image and object are at the same distance from the lens 
they will be of the same size and the image will increase in size 



Theoretic Optics 25 

in proportion to its distance. Therefore in order to obtain a 
real image four times the size of the object its distance must be 
increased four times. 

In this case the lens is a + 2.50 D. with a focal distance of 
16 inches of which one-fifth w411 represent distance of image and 
four-fifths distance of the object. This is 1/80 and 1/20 re- 
spectively, or 80 inches for the distance of the image and 20 inches 
for distance of object. This is proven to be correct by the 
transposition of these inch numbers into diopters 80 inches = .50 
D. and 20 inches = 2 D., the sum of which is 2.50 D., the number 
of the lens given in the question. 



How is the curvature of a spherical wave of light measured? 
How would you verify the fact that when a spherical wave of light 
passes directly through a thin lens the change in the curvature of 
the wave is the same for all positions of the lens? 

The curvature of the wave front is inversely proportional to 
the distance of its source, and hence the curvature of a spherical 
wave of light is measured by the reciprocal of that distance. 
If a one meter length of radius be taken as the unit, the curvature 
of the wave front at a distance of one meter is said to be 1 D. 
At half a meter it is 2 D., at one- third meter 3D., and at two 
meters .50 D. 

Parallel rays of light or a plane wave, passing through a 
1 D. lens, will be given a curvature of 1 D., and so on. 

In order to prove that the change in curvature of a spherical 
wave in passing through a thin lens is constant, we fall back 
upon the regular formula for conjugate foci: 

0^ i f 

in which o represents the distance of the object, i the distance of 
the image and / the principal focus. 

— expresses the curvature of the wave diverging from the 
object and meeting the lens, ^ the curvature of those leaving the 

lens to form the image and -j the focal length of the lens. The 

. 1 . . ... 1 

latter must be a constant and if — is increased or diminished, -^ 

must be decreased or increased in the same ratio. 



26 State Board Examinations 

If the curve of a biconvex lens on one surface is on a radius of 
13}^ inches and on the other on a 40-inch radius, what is the power 
of the lens? 

Such a lens is equivalent to one plano-convex lens with a 
radius of 3 D. and another plano-convex lens of 1 D. radius. 
Now, then, the focal length of a plano-convex lens is equal to 
twice the length of its radius, or what is equivalent to the same 
thing one-half its refractive power. Applying this rule we find 
the combined power of the two surfaces is 4 D., one-half of 
which is 2 D., and this represents the power of the lens. 



In what three ways may light he affected? 

It may be refracted, reflected or absorbed. Or its velocity 
may be increased or diminished, or bent out of its course. 



What is an image and how is it formed? 

An image is the exact reproduction of an object, and is 
formed by refraction through a lens or reflection from a mirror. 

There are two kinds of images, real and virtual. The first 
is formed by a meeting of the rays, it has an actual existence and 
may be projected upon a screen. The second is only an imag- 
inary image such as would be formed by a prolongation back- 
ward of the diverging rays of light. 



What is a ray of light called before passing from one medium 
to another? 

An incident ray. 



What is a ray of light called after it has passed from one 
medium into another? 

A refracted ray. 



What is spherical aberration? 

Spherical aberration is the wandering of rays from a single 



Theoretic Optics 27 

focus, or the focusing of the rays of Hght passing through a lens 
at varying distances from the lens, the rays passing through 
the periphery coming to a sooner focus than the more central 
rays. This is due to the fact that the curvature and therefore 
the refractive power of the lens increases from the center (where 
it is nothing) to the periphery (where it is greatest). 



What is chromatic aberration? 

Chromatic aberration is due to dispersion of light on account 
of the difference in refraction of the different colors of which 
white light is composed. As a result in the image formed by an 
ordinary lens, the outlines of the figures will be found to be 
edged with rainbow hues. The seven primary colors each have 
a different degree of refrangibility, the red being least refracted, 
the violet most, the other colors in regular order between these 
extremes. 

If light is reflected away from the normal or perpendicular, 
what is the cause? If toward the perpendicular? 

In the first case on account of the passage of rays from a 
dense to a rare medium; and in the second case the passage of 
lizht from a rare to a dense medium. 



What is a virtual focus? 

This is synonymous with a negative focus, and is the point 
from which rays appear to diverge after passing through a 
concave lens, or after reflection from a convex mirror, or after 
refraction by convex lens when object is closer than its principal 
focal distance, or after reflection from a concave mirror when 
object is closer than its principal focal distance. 



What is reflection of light? 

This is the throwing back of rays of light from the surface 
on which they fall into the same medium through which they 
came. The best form of a reflector is a mirror or a polished metal 
surface. But even if the body is transparent and the rays are 



28 State Board Examinations 

transmitted, yet some of them are reflected and it is by these 
reflected rays that objects are made visible. 



What is refraction of light? 

The deviation which takes place in the direction of a ray 
of light as it passes from one medium into another of different 
density. The two laws that govern refraction are: 

1. In passing from rare to dense medium, bent toward the 
perpendicular. 

2. In passing from dense to rare medium, bent from the 
perpendicular. 

Where is the principal focus of a lens? 

At the principal focal distance of a lens, that is where 
parallel rays are made to meet after being refracted by a convex 
lens. If the rays that enter the lens are convergent or divergent 
instead of parallel, the focus would be shortened or lengthened, 
but it would not be the principal focus. 



What are conjugate foci? 

The point from which rays originate and diverge and the 
point on the other side of the lens on the axial ray w^here they 
come together are conjugate to each other, and are known as 
conjugate foci. Either one may be considered as the object and 
the other will be the image. 



What is the difference in the image as formed by a concave 
spherical mirror and a convex spherical mirror? 

Assuming the object to be at a distance so that the rays that 
strike the mirror are parallel, the image formed by a concave 
mirror is real, inverted, and in front of the mirror and can be 
caught upon a screen ; while that formed by a convex mirror will 
be virtual, erect, behind the mirror and cannot be caught upon 
a screen. 



Theoretic Optics 29 

What is the conjugate focus to a point which is 2 inches from 
a minus lens with a principal focus of 3 inches? 

If the source of light is at a distance of 2 inches the rays of 
Hght would have a divergence equal to — 20 D., which will be 
increased to ?>?> D. by passing through a concave lens with a 
divergence of 13 D., as shown by the principal focus of 3 inches. 
This ?)?) D. will show a virtual focus at about 1 1/5 inches on the 
same side of the lens as the object is situated; and therefore 
11/5 inches would be conjugate to 2 inches in this case. 



A plus lens has a principal focus of 40 cm. and the radius 
of curvature of one of its surfaces is 40 cm.; what is the radius 
of curvature of the other surface, the index of refraction of the glass 
being 1.5? 

In a biconvex lens the focal length is just equal to its radius; 
therefore in this case, where the principal focus is identical with 
the radius of curvature of one of its surfaces, the lens must be 
biconvex and the radius of the second surface the same as that 
of the first, viz., 40 cm. 



If the index of refraction of a certain kind of glass is 1.496 
on what radius of curvature must a plano-convex lens he struck so 
that the lens will he a 1 D. lens? 

The rule is, focal distance, multiplied by index of refraction 
equals radius of curvature if the lens is piano. Substituting 
figures, we have 

40 (1 D.) X (1.496 - 1) 

or 40 X .496 = 19.84 

The radius of curvature would have to be 19.84 inches. 



What is the critical angle in refraction and what is the result 
of same? 

This is the least angle of incidence that permits a ray of 
light traveling in a dense medium to pass into a rare medium', 
and beyond which total reflection occurs at the surface which 
separates the two media. As this prevents any loss of light from 
transmission or absorption, the reflection will be the most brilliant 



30 State Board Examinations 

obtainable, and hence this method is made use of in optical 
instruments. 

The critical angle from water to air is 48° 35', which means 
that this is the least angle that will permit the light to be refracted 
and pass out of the water. If the rays form an angle greater than 
48° 35', they will not pass out of the water but will be reflected 
back into it. The surface separating the two media becomes a 
reflecting surface and acts as a plane mirror. 



What is the conjugate focus to a point 36 centimeters from a 
biconvex lens of 2.5 D.? 

A convex lens of 2.50 has a principal focal distance of 40 cm., 
and rays proceeding from this point would emerge from the lens 
parallel. If they proceed from a point farther than the principal 
focus they will emerge from the lens convergent and will meet at 
a point which is conjugate to that from which it proceeded. On 
the other hand, if the rays proceed from a point closer than the 
principal focus, they will emerge from the lens as divergent and 
will not meet in a focus. 

The latter is the condition that applies in this question, the 
principal focus being at 40 cm. and the rays proceeding from a 
point at 36 cm. The rays will emerge still divergent, but not so 
much so as before entering the lens. If these divergent rays 
were continued backward by imaginary lines to the point from 
which they appear to proceed, that point would be conjugate to 
the first, although it is a negative focus and has no real existence. 

The question is, where will that point be located in the case 
under consideration? Rays proceeding from 36 cm. will have a 
divergence equal to 2.85 D. After passing through the + 2.5 D. 
lens and being refracted to that extent, they will emerge with a 
divergence equal to .35 D., which represents a distance of 114 
inches, or 285 cm. which in this case is conjugate to the 36 cm. 



What general effect do plus and minus lenses have on diverging 
and converging light? 

Convex lenses act on diverging rays so as to lessen their 
divergence. If the rays proceed from some distance or from a 
point beyond the principal focus, the divergence will be entirely 



Theoretic Optics 31 

overcome and the rays made convergent and brought to a focus. 
If proceeding from its principal focus, the convex lens would 
make them parallel. If proceeding from a point closer than the 
principal focus, the convex lens would reduce their divergence. 

The convex lenses act on converging rays in such a way as 
to make them more convergent and bring them to a sooner focus. 

Concave lenses act on divergent rays in such a way as to 
increase their divergence. 

Concave lenses act on convergent rays to lessen their con- 
vergence. If their divergence corresponded with the lens they 
would emerge parallel; otherwise they would meet in front of 
or beyond the negative focus of the lens, depending upon the 
degree of their divergence. 

A certain lens focuses parallel light at 20 cm.; what lens 
must he placed in front of it to send the focus hack to one meter? 

The lens in question must be convex and if it focuses parallel 
rays at 20 cm. this must be its principal focal distance, in which 
case the lens would be + 5 D. 

In order to throw the principal focus back to one meter, 
the strength of the lens must be reduced to + 1 D. which can be 
accomplished by placing in front of the original lens a — 4 D. 



What is the strength of a lens that causes light coming from a 
point 133 cm. away to pass out parallel? 

If light emerges from a lens in parallel lines it must come 
from a point at the principal focus of the lens, because the latter 
is conjugate with infinity, as represented by the parallel lines. 
Therefore the strength of the lens is approximately .75 D,, which 
is found by dividing 133 cm. into 1 meter (or 100 cm.). 



What is the index of refraction when light passes from water 
into glass, the latter having a refractive index of 1.546? 

The law of refraction is expressed by the following formula : 

Sine angle incidence . , , ^ 

^. —. -, — -:— = index of refraction. 

Sine angle refraction 

This constant quantity, which is known as index of refrac- 
tion, represents the relative velocity of light in the two media, the 



32 State Board Examinations 

greater the density the less the velocity. The sine of the angle 
of incidence always bears the same numerical relation to the sine 
of the angle of refraction, no matter what the angle of incidence 
may be. 

The index for two media with reference to each other is 
obtained by dividing the refractive index of one into the refrac- 
tive index of the other. 

In this question we have the refractive index of glass given 
as 1.546, which we divide by the refractive index of water (1.333), 

1.333)1.546(1.1597 
1.333 
2.130 
1.333 



7.970 
6.665 

1.3050 

1.1997 

1.0530 

.9331 



A certain lens in air has a power of 6 diopters, the refractive 
index of the glass being 1.562; what will he the dioptric power of 
this lens when submerged in water? 

In order to obtain the refractive index for water and glass 
with reference to each other, w^e divide the index of glass (1.562) 
by the index of water (1.333), and the result is 1.171. 

In order therefore to compare the power of the lens in water 
with its power in air, we must make use of the proportion of the 
refractive index of one to the other. 

X : 6 D : : .171 : .562 

X = 6D x47T-= 1-82 D. 
.562 

Therefore the power of this 6 D. lens when submerged in 
water is reduced to 1.82 D. 



A plus 8 D. lens made of glass of refractive index of 1.523 is 
cemented inside of two plates ground with curves to match its surfaces, 
hut of index of refraction of 1.634; what is the power of the lens 
in its new position? 



Theoretic Optics - ZZ 

In order to obtain the refractive index of these two kinds 
of glass with reference to each other, w^e divide the index of one 
(1.523) into the index of the other (1.634) and the result is 1.073, 
which is the refractive index of the lens as it is enclosed in the 
two plates of glass. 

In order to ascertain its power we have the following pro- 
portion : 

X : 8 D : : .073 : .523 
X = 8 X .073 = 1.11 D. 
.523 

Therefore the power of this 8 D. lens inside the enclosing 
plates of glass has been reduced to 1.11 D. 



An 8 D. lens is decentered 4.5 mm.; what is the prismatic 
effect of the deviation in prism diopters? 

The unit of measurement is 1 p. d. for each diopter of 
refractive power, when decentered 10 mm. 

An 8 D. lens decentered 10 mm. = 8 p. d. 
" " " '' 1 mm. = .8 p. d. 
" " " " 4.50 mm. = 3.6 p. d. 
The prismatic effect of an 8 D. lens decentered 4.5 mm. is 
3.6 p. d. 

An object is placed 9 inches in froiit of a lens and an image of 
this object is formed at 18 inches from the lens on the other side; 
what is the principal focus of the lens in inches? Make the calcula- 
tion. 

^ ^- / 



or substituting figures 

The principal focus of this lens is 6 inches. 



9 + 18 ~ 18 °^ 6 



How are rays of light reflected from piano, concave ana convex 
mirrors? 

No matter whether the surface is piano, concave or convex, 
the laws of reflection are always the same, viz. : 



34 State Board Examinations 

The angles of reflection and incidence are equal. 

The incident and reflected rays are always in a plane per- 
pendicular to the reflecting surface. 

In the case of a plane mirror the incident ray is reflected in 
the opposite direction at the same angle. 

In the case of a concave mirror the rays are reflected accord- 
ing to the same law, and if parallel are converged to a point 
which is the principal focus of the mirror, and is approximately 
half its radius of curvature. 

In the case of a convex mirror the rays of light according 
to the same law are reflected divergently. 



The index of a certain substance is stated to he 1.564; what 
is the meaning of this statement? 

By the index of refraction of a substance is meant its relative 
density or the comparative length of time required for light to 
travel a definite distance in different substances. The greater 
the density the slower the velocity, and the smaller the angle of 
refraction. 

The index of refraction is found by dividing the sine of the 
angle of incidence by the sine of the angle of refraction. In 
this particular case the result was 1.564 which has been deter- 
mined with reference to air, the index of which is regarded as 
unity. Or, in other words, the lessened rapidity with which 
light passes through this substance as compared with air, the 
ratio being 1 to 1.564. 

When a ray of light passes from a denser to a lighter medium, 
which way is the ray bent? 

According to the laws of refraction in passing from a dense 
to a rare medium, as from glass into air, if the ray is oblique it 
is bent from the perpendicular. If the ray impinges upon the 
second medium at right angles it passes unrefracted. 



An object is 24 inches high and it is 60 inches from a lens; 
the image of this object is 10 inches from the lens; what is the height 
of the image? 



Theoretic Optics 35 

If i is size of image, o size of object, d i distance of image, 
and d o distance of object, the proportion is expressed as follows: 

i : :: d i : d o 
Substituting the figures in the above question we have 
Z : 24 :: 10 : 60 

V 240 , . , 
X = -77- = 4 inches. 
oU 



An image formed by a certain lens is 5 mm. wide and at a distance 
of 200 mm. from the lens. The object is 4 meters distant; how wide 
is it? 

Using the same proportion and substituting figures we have 
5 mm. : X :: 200 mm. : 4 m. or 4000 mm. 

^ 20000 .„„ 

^ = "200- = ^^^ "^^ 

This is equivalent to 10 cm. or l/lO meter. 



The distance of an object from a certain lens is 24 feet {288) 
inches; the distance of the image of this object from the lens is 6 inches 
on the same side of the lens as the object; what is the principal 
focus of the lens? 

If we assumed that the rays from 20 feet and beyond were 
parallel then these rays from 24 feet forming an image at 6 inches 
would indicate this distance as the principal focus, and as the 
image is on the same side of the lens as the object, it must be 
virtual and the lens concave. 

To work it out accurately we use this general formula: 

o ^ ^" / 

in which represents the distance of the object, i the distance 
of the image, and / the principal focus of the lens. 
Substituting figures we have 

2-88 +"6 = 288°^^-^^ ^""^"'- 

In this case the focus is .12 or 1/8 inch closer than if we assumed 
the rays to be parallel. 



36 State Board Examinations 

The principal focus of a certain plus lens is 10 inches; the 
object is placed 20 inches from the lens; where will the image of the 
object be found? 

10 20 20 
Image of object will be found at 20 inches. 



An object is ^ meter from a -\~ 6 D. lens; where will the 
image be? 

Using the regular formula and substituting values, 
— = 6 D. - 2 D. = 4 D. 



1 1 

— =—r meter. 
4 

The image will be yi meter from the lens. 



A certain plano-convex lens has a principal focus of 40 mm. 
and the index of refraction of the glass is 1.620; what is the radius 
of curvature of the convex surface? 

The rule is that the focus multiplied by index of refraction 
less unity equals radius of curvature. 
Substituting values 

40 (1.620 - 1) = 40 X .620 = 24.8. 
The radius of curvature of the convex surface is 24.8 mm. 



A certain lens has an index of 1.500, one surface being on a 
radius of 12 inches and the other on a radius of 24 inches; what 
will be the principal focus of the lens? 

Inasmuch as the refractive index is exactly 1.50, and no 
more, we may say that the focal length of a plano-convex lens is 
equal to twice the length of its radius. Therefore, the focal 
length of the 12-inch surface is 24 inches, and of the 24-inch 
surface, 48 inches. 

Then applying our formula we have 

-4-1- =-1 1 

24 + 48 "" 48 ' °^ 16 

That is, the principal focus is 16 inches from the lens. 



Theoretic Optics 37 

In transposition of tens power does the cylinder ever change 
in value? What is the reason? 

In transposition of sphero-cylindrical lenses the number of 
the cylinder must always remain the same, although there Is a 
change In Its sign and In the position of Its axis. The amount 
of the cylinder cannot be changed because It represents the degree 
of astigmatism or the difference In power between the meridians 
of least and greatest curvature. This Is an Important point for 
the student to remember, because If he loses sight of It he soon 
falls Into error In working out problems in transposition. 



How is it possible to measure the strength of a plus lens without 
using either neutralizing lenses or a lens measure? 

By measuring Its focal distance. By allowing the rays from 
an object at least 20 feet distant, as a tree, or a house, or a sign, 
to pass through the lens and form an Image on a screen. The 
lens is slowly moved closer to and farther from the screen until 
the image is found to be the most distinct, and then the distance 
of the lens from the screen will represent the principal focal 
distance of the lens, which can then be converted Into diopters 
to show Its refractive power. 



Object is 14 inches high and its distance from a 2-inch focus 
concave mirror is 40 inches; what is the height of the image? 

First w^e must find the distance of the image. If distance of 
object is 40 inches from a 2-inch concave mirror the distance of 
image is found by the following formula: 

_1 1 _ 19 

2 40 ~ 40 

That Is, Image Is at 2 2/19 Inches. 

Now then, In order to find the height of Image, we make use 
of the usual proportion 

X = 14/19 of an Inch, which is the height of the Image. 



38 State Board Examinations 

Object is 80 inches high, its image is 2 inches high, the mirror 
being convex. The distance of the object from the mirror is 80 
inches; what is the principal focus of the mirror? 

If an object is 80 inches high and its distance is 80 inches, 
then if the image is 2 inches high its distance will be 2 inches. 

Having the distance of the object (80 inches) and the distance 
of the image (2 inches), the principal focus of the mirror is found 
by the following formula: 

11 39 2 . , 



A 3 -inch concave mirror forms an image of a certain object, 
the object lyi feet high and the image 1}4 inches high. How far 
away is the object from the mirror? 

We have the size of object and the size of image, and in 
order to find distance of object, we must work out the distance 
of image as the third proportion. 

The object is twelve times the size of the image, therefore 
the distance of object must be twelve times that of image. The 
whole focal distance is 3 inches, of which ^"^As represents distance 
of image and ^is distance of object. 

-1 f ^ = i- 
13 ° 3 39 

that is, 39 inches is the distance of object from the mirror. 

This can be proven by finding the distance of image as ■'^%9, 
and then using formula: 



1 , 12 13 1 
or 3-inch focus. 



39 ^ 39 39 °^ 3 



A n object which is 2 feet wide is at a distance of 8 feet from a 
4 D. convex mirror; what is the width of the image? 

In a convex mirror the focus is negative, but otherwise the 
rules governing the calculations of conjugate foci are the same 
as with a concave mirror. 



Theoretic Optics 39 

In this case we have the distance of the object and the 
principal focal distance of the mirror, and we get the distance 
of the image by the following formula: 

_ J_ _ J^_ 1_ 

10 96 3 

9— 
53 

The image is virtual and 9%3 inches behind the mirror. 

Now then we find the width of the image by the following 

proportion : 

96 : 24 : : 9 3/53 : X 
Or as the size of the object is % of its distance, so the size 
of the image will be >^ of its distance. 

X = 34 of 9 3/53 = 2 14/53 inches 
That is, the image will have a width of 2Ho3 inches. 



An object which is 12 inches in diameter is 40 feet from a 
concave mirror. The image of the object formed by the mirror 
is 4 inches in diameter. On what radius is the mirror surface 
formed? 

In the first place we find the distance of the image by the 
following proportion: 

Size Object Distance Object Size Image Distance" Image 
12 in. : 480 in. : : 4 in. : X 

^ 480X4 _^. , 
X = -^ = 160 inches 

Having the distance of the object and the distance of the 
image, we find the principal focal distance by the following 
formula: 

o ^ i f 



Substituting figures 



_J_ J_ ^_4 J_ 
480 ^ 160 480 °^ 120 



120 inches, or 10 feet is the focus of the mirror. 

And as the radius of curvature of a concave mirror is twice 
the principal focus, in this case the radius on which the mirror 
surface is formed must be 20 feet. 



40 State Board Examinations 

An object is 20 feet from a convex mirror. Its height is six 
times as great as the height of its image; what is the principal 
focus of the mirror? 

If the height of the image is 1 6 the height of the object, 
then the distance of the image must be 1/6 the distance of the 
object; this would give us 40 inches as the distance of the image. 
The distance of the object being 240 inches and of the image 
40 inches, we find the principal focus of the mirror by the follow- 
ing formula, always remembering that in a convex mirror the 
focus is negative. 1 1 _ 5 _ 1 

40 ~ 240 ~ l40 ~ 48" 

48 inches is the principal focus of the convex mirror. 



What is the radius of curvature of the second surface of a 
lens, of which the first surface has a radius of curvature of 10 cm., 
the principal focus of the entire lens being 25 cm., and the index 
of refraction being 1.5? 

If the radius of curvature of the first surface of the lens is 
10 cm., its focal length would be twice that of the radius, or 20 
cm., which is equivalent to a dioptric power of -f 5 D. The 
principal focus of the entire lens is 25 cm., which is equivalent 
to + 4 D. Hence the dioptric power of the second surface of 
the lens must be — 1 D., which would correspond to a focal 
length of 100 cm. and a radius of 50 cm. 



An object 3 inches in diameter is placed in front of a concave 
mirror at such a point that its image is formed 4 inches from the 
mirror, the diameter of the images being ^ inch; what is the focus 
of the mirror? 

In this example the distance of the image is 4 inches and its 
size }i inch ; that is, the distance is eight times the size, therefore 
the distance of the object would be eight times its diameter; 
the latter being 3 inches, the former would be 24 inches. 

Then we find the focus of the mirror by the usual formula: 

1^1 7 1 

1 + 1 = 4 or 4- + 24=T4-J 

o 1 f 3- 

7 
3 3/7 inches is the focal distance of the mirror. 



Theoretic Optics 41 

Transpose the following to two forms of sphero-cylinders: 
+ 1 cyl. ax. 90° with 1 cyl. ax. 180°. 

4- 1 D. sph. C - 2 D. cyl. ax. 180°, and 
- 1 D. sph. C + 2 D. cyl. ax. 90° 



When a ray of light passes through a denser medium than 
air — a piece of glass for instance — and it is apparently bent or 
refracted from a straight line, why does it not continue in the same 
course after emerging from the glass that it assumes while passing 
through the glass? Why does it change back to the same angle as 
before entering the glass? 

Because it is subject to the laws of refraction, and hence 
in passing from a rare to a dense medium, as from air into the 
piece of glass, it is bent toward the perpendicular; and in passing 
from a dense to a rare medium, as when it emerges from the 
glass into air, it is bent from the perpendicular. The amount of 
bending in each case is equal, and hence the emergent ray is 
parallel to the entering ray. 

A certain lens has one surface convex on a radius of four 
inches, and the other surface coitvex on a radius of ten inches; index 
of refraction 1.60; what is the power of the lens? 

The focal distance of each surface is found by dividing the 
radius of curvature by the index of refraction, less one. 

4. 4. 

= ^ =6 2/3 = 6D. 



(1.60 - 1) .60 
10 10 



(1.60 - 1) .60 



= 16 2/3 = 2.50D 



6 D. + 2.50 D. = + 8.50 D. approximately the power of the 
lens. 



Why is it that as a lens is moved from side to side a distant 
object seen through the lens seems to have a movement of its own? 

Because at every point except at its optical center the 
lens will show a prismatic effect, and because the effect of a 
prism is to cause displacement in the direction of its apex. In a 



42 State Board Examinations 

lens the prismatic effect increases from the center to the periphery, 
and hence as a lens is moved from side to side and the visual 
line passes through the lens at different points from center to 
periphery' the prismatic power of the lens comes into evidence 
and causes displacement. As the lens is kept moving the dis- 
placement shows itself as a movement of the object, which in 
concave lenses is in the same direction, and in convex lenses 
opposite. 

In what direction relative to each other do rays of light travel 
when first leaving a luminous point? 

Divergently. 



What effect do the following have upon parallel light: Convex 
sphere, concave sphere, plano-convex cylinder? 

Convex sphere converges to a focal point; concave sphere 
diverges as from a focal point, and convex cylinder converges to 
a focal line. 

What three things happen to a ray of light incident on a glass 
surface? Refraction {if the surface is curved), reflection and 
absorption. 

State a rule for the transposition of a compound to its equivalent 
form which will apply to both minus and plus forms of compounds. 

The new sphere is obtained by the algebraic addition of the 
sphere and cylinder, and the new cylinder retains the number 
of the old one but changes its sign and axis. 



What is the dioptric power of a plus lens with a focus of 984 
inches, the index of refraction being 1.742? 

In a case w^here the radius of curvature is given we must 
know the index of refraction in order to figure out the focal 
distance of the lens. But in a question like this, where the 
focal distance is given with a desire to transpose it into diopters, 
the index of refraction is not taken into account, but we simply 



Theoretic Optics 43 

divide the inch number into 40 and the result will be .04 D., or 
approximately 1/25 of a diopter. 



What is the radius of curvature for the two surfaces of a + 3D. 
lens with a — 9 D. surface on the inner side, index 1.57? 

If the concave surface of this + 3 D. lens is — 9 D. the 
convex surface is + 12 D. 

In this case where the index is 1.57 we solve the problem as 
follows : 

3.33 X (1.57 - 1) = 1.89 inches ' 
4.50 X (1.57 - 1) = 2.56 inches 



To find the radius of curvature for the surface of a lens we 
are told to deduct unity from the index of refraction; what does 
this mean, and give an example that will apply in practice? 

If the index of refraction is 1.54 and we deduct unity, there 
would be left .54, which is the excess above unity and the amount 
in which we are interested. 

Suppose the focal length of the lens was ten inches, the 
problem would be focal length multiplied by index of refraction 
less unity equals radius of curvature, or 10 X (1.54 — 1) = 10 X 
.54 = 5.4. In this case if the focal length is ten inches and the 
index of refraction 1.54 the radius of curvature is 5.4 inches. 



What is the fundamental reason for light being refracted as it 
passes from one substance into another? 

Because of the difference in the density or index of refraction 
of the two substances. If light would pass from one substance 
into another, both of the same density, there would be no refrac- 
tion. As the density of a substance increases, the velocity of 
light is diminished and then refraction takes place. 



Find the radius of curvature of a lens for its two surfaces, 
each surface to be alike and each to measure one diopter. Index of 
glass, 1.632. 



44 State Board Examinations 

Multiply the focal length by twice the index of refraction 
less one. In this case where the two surfaces each measure one 
diopter the focal length of the lens is twenty inches. 
The problem is 

20 X 2 (1.632 - 1) 
or, 

20 X 1.264 = 25.28 inches, 
which is the radius of curvature. 



How would you transpose the following cross-cylinder into an 
equivalent sphero-cylinder without using a mathematical formula 
for the same: + .75 cyl. axis 120° combined with — 2 cyl. axis 60°? 

I would take these two cylinders from the trial case and 
place them in the positions indicated in the trial frame. Then I 
would look through this combination at a straight line and 
locate the two principal meridians, which I would then proceed 
to neutralize with convex or concave spheres. Or, in other 
words, inasmuch as this obliquely crossed cylinder is equivalent 
to a sphero-cylinder, I would neutralize the combination in the 
same way that any sphero-cylinder would be neutralized. Then 
the opposite of this neutralizing combination would be the 
transposition desired. 

With a candle flame one meter from two -^ 3 D. lenses placed 
13 inches apart, where should the screen he placed to receive a perfect 
image of the candle flame? 

The divergent rays from the candle 40 inches away after 
being refracted by the first lens are converged to a point 20 
inches on the other side of the lens, but on their way they meet 
the second lens placed 13 inches from the first, and they are 
then convergent to a point 7 back of the second lens (20 — 13 = 
7). The effectivity then of the first lens in the plane of the second 
lens is that of 1/7, or the effect is the same as if a 7-inch lens was 
placed in contact with the second lens, and the screen to receive 
a perfect image of the flame would be placed at 4 11/20 inches 
being found as follows: 

1 , 1 20 , 11 . , 

y +13 9i°^S-o-^"^^^^ 



Theoretic Optics 



45 



If Fr = resultant focus, 
If F] = focus of first lens, 
If F2 = focus of second lens. ~ 

If d = distance between the two lenses, then the effective focal 
distance is found by the following formula: 

(Fi - d) ¥, 



Fr = 

Substituting figures we have 

(20 - 13) 13 



Fr 



20 



13 



13 



Fi +F2 



7 X 13 
33 - 13 



-^ = 4 1^ inches 



What is the rule to figure the curvatures of the surfaces of lenses? 



/^ E 


1.2^3^4>\^ 




will 


w 


G^--^_ 


^ijulu^ 



Fig. 9. (See next page) 



Focus multiplied by index of refraction less unity = radius 
of curvature. 

For instance, take a 20-inch lens of an index of refraction 
of 1.62. 

20 (1.62 - 1) = 12 4/10 inches. 

Or if we wished to use the metric system we would say that 
with an index of refraction of 1.62 the radius of curvature for a 
1 D. surface is 62 cm. and for a 2 D. surface 31 cm., which 
corresponds to the answer of 12 4/10 inches found above. 



46 State Board Examinations 

What happens when light strikes a polished opaque surface 
and a polished transparent surface? 

When light falls upon a polished opaque surface it is prac- 
tically all reflected. When it falls upon a polished transparent 
surface it is nearly all transmitted, although there is some 
reflection. 

What is meant hy index of refraction? Give example to find 
relative index of water. 

Commonly we understand it to be the refractive or bending 
power of a medium as compared with air, which is standard, and 
the index of which is unity. The index of refraction depends 
upon the relative density of a substance or the comparative 
length of time required for light to travel a definite distance 
in different substances. The greater the density the slower the 
velocity and the more the refractive power. 

In order to find the index of refraction of any substance as 
compared with air we divide the sine of the angle of incidence 
by the sine of the angle of refraction. (See Fig. 9 on preceding 
page.) 

Let A Bh^ the refracting surface separating air from water. 
Let C jD be a ray incident on the surface at the point D, to which 
the line E D F is the perpendicular. The C D E is the angle of 
incidence and G D F is the angle of refraction. Then £ C is 
the sine of the angle of incidence and G F the sine of the angle 
of refraction. 

We divide the sine of the angle of incidence (4) by the sine 
of the angle of refraction (3), and the answer is 4/3, or 1 1/3, or 
L33, which is the index of refraction of water as compared with 
air. 

When does a real image produced hy a plus lens change to a 
negative image? 

When the object is closer to the lens than its principal focal 
distance the rays will emerge divergently and the image will be 
a negative one. 

What is the rule for the transposition of a sphero-cylinder 
into its equivalent form? 



Theoretic Optics 47 

The new sphere is obtained by the algebraic addition of the 
sphere and cyHnder of the old combination. 

The new cylinder retains the vsame number or dioptric value 
as the old one, but there is a change of sign and axis. 



If a + 4 D. lens is so placed in relation to light that it forms 
an image of the light 100 feet away, what is the relative size of object 
and image? 

If the image is formed at a distance of 1,200 inches from a 
ten-inch lens, then the distance of the object is found by the 

following equation: 

1 _ _1_ _ 119^ 
10 '1200 ~ 1200 

If the distances of image and object are as 1 to 119, the 
relative sizes will be in the same proportion. 



What is the radius of curvature of a double convex lens of 5 D. 
of power, the refractive index of the glass being 1.542? 

When glass has a refractive index of 1.50 the radius of 
curvature of a biconvex lens and its focal distance are equal. • 

But in this case where the index is 1.542 the radius of curva- 
ture of this eight-inch focus lens is 8^^%ooo inches, as found by the 
following formula : 

8X2 (1.542 - 1) = 8 X 1.084 = 8.672 



What is the refractive index of a piano lens of 2 D., the radius 
of curvature being 25 cm.? How do you make the calculation? 

We divide the focus into the radius, as 25 cm. divided by 
50 cm. (the focal length of a 2 D. lens) equals .50. This being 
the amount above unity, the index of refraction is 1.50. 



What is the focal length of two lenses placed in contact whose 
powers are -\- 3 D. and -f 5 D.? 

The two lenses would be equivalent to a -|- 8 D. lens with a 
focal length of five inches. 



48 State Board Examinations 

Transpose the following so as to eliminate spheres: 

(a) -\- 5 D. sphere = - 4 D. cyl. axis 10° 

(b) + 4 D. sphere = - 4 D. cyl. axis 120° 

(a) In this case we can eliminate the sphere only by trans- 
posing into a cross-cylinder, as follows: 

+ 1 D. cyl. axis 10° = + 5 D. cyl. axis 100° 

(b) In this case, where the sphere and cylinder have the 
same value, the sphere can be eliminated by transposition into 
a piano cylinder, as follows: 

-f 4 D. cvl. axis 30° 



Transpose the following sphere cylinder to its equivalent form 
explain how you do it: 

+ 1 D. sphere = - .50 D. cyl. axis 120° 

The new sphere is obtained by the algebraic addition of 
the sphere and cylinder, as — .50 added to + 1 = -f .50 D., which 
is the new sphere. 

The new cylinder has the same value as the old one, but 
its sign and axis changes from — .50 D. to + .50 D. and from 
axis 120° to axis 30°. 

The completed transposition is: -f .50 D. = + .50 D. cyl. 
axis 30°. 



Define (a) a ray of light; (b) a pencil of 

(a) The smallest subdivision of light traveling in a straight 
line and perpendicular to a wave front. 

(b) A bundle of light, cone-shaped and composed of con- 
vergent or divergent rays. 



When light falls upon a transparent plate, what is the incident 
ray and which is the emergent ray? 

The incident ray is that which strikes the plate after leaving 
the luminous point. The emergent ray is that which leaves the 
plate after having passed through it. 



What three things may happen to light when it falls on a surface? 



Theoretic Optics 49 

Reflected, absorbed and transmitted or refracted. 



A re rays of light coming from an object plus or minus? 

Rays of light coming from an object are divergent and 
might be considered as minus, because they require a convex 
lens to overcome the divergence and make them parallel. 



What are the conjugate foci of a lens? 

The location of object and image which are interchangeable. 



What is a real focus and what is a virtual focus? Give an 
example of each. 

A real focus is formed by the actual meeting of rays after 
refraction by a convex lens and can be caught on a screen. 

A virtual focus is formed by the imaginary continuation 
backward of divergent rays, as in the case of a convex mirror. 



At what distance from the surface of the mirror will parallel 
light he focused, if the radius of curvature of the mirror is 20 inches? 

The principal focal distance of a concave mirror is approx- 
imately one-half the radius. Therefore, in this case, where the 
radius is 20 inches, parallel rays would be focused at 10 inches. 



Transpose the following : 

(a) - 3.25 C: -\- 1.50 cyl. axis 45° 
{b) -f .75 O - .75 cyl. axis 90° 

(a) - 1.75 sph. = - 1.50 cyl. axis 135° 

(b) + .75 cyl. axis 180° 



What is the combined value of -\- 3, — 1, + 4, — 6? 
Zero. ^ 

From — 2.50 take away + 3.25 
- 5.75. 



50 State Board Examinations 

Change + .12 C^ — .25 cyl. axis 50° to its equivalent sphero- 
cylinder. 

- .12 C + .25 cyl. axis 140°. 



Change the following to its equivalent peris copic form: 
-h 1.25 sph. O + 1.25 cyl. axis 120° 

+ 2.50 D. sph. C - 1.25 D. cyl. axis 30° 



Transpose to corresponding sphero-cylinders: 

(a) - .50 O + .75 cyl. axis 15° 

(b) + .50 cyl. axis 50° 

(c) -\- .50 :2 - .75 cyl. axis 45° 

(d) - .50 cyl. axis 90° 

(a) + .25 D. sph. = - .75 D. cyl. axis 105' 

(b) + .50 sph. C - .50 cyl. axis 140° 

(c) - .25 sph. C + .75 cyl. axis 135° 

(d) - .50 sph. C + .50 cyl. axis 180° 



Does the speed of light hear any relation to the index of refrac- 
tion, and if so, what? 

The velocity of light is lessened in media denser than air, 
the decrease in speed being roughly proportionate to the density. 
The velocity of light in the first medium is to the velocity of 
light in the second medium as the index of refraction of the 
second medium is to the index of refraction of the first medium. 
Or, in other words, the rate of progression of light in a medium 
is inversely proportional to its optical density. For example, 
if the index of refraction of glass was 1.50 or 3/2, then the speed 
of light as it passes from air into this glass is reduced to 2/3. 



When light falls vertically upon a plane glass surface, what 
change is produced, if any? 

The light would pass unrefracted, the only change being 
that its speed would be reduced to two-thirds approximately. 



How can the existence of refraction be shown with a penny 
and a cup of water? 



Theoretic Optics 51 

The penny is placed at the bottom of the cup, with the eye 
of the observer in such a position that it cannot be seen. The 
cup is filled with water, and without any change in the position 
of the eye, the penny comes into view. 



In what way does a light wave differ from a ray of light? 

In studying the transmission of light, there is assumed to 
be a disturbance in the ether which exists throughout the uni- 
verse, manifesting itself by a series of waves, which is usually 
likened to the familiar example afforded by throwing a stone 
into a pool of still water. Therefore the wave has a circular or 
spherical front. 

A ray of light is the smallest conceivable line of light, and 
may be regarded as a straight line perpendicular to the wave 
front. 

// we place a transparent solid in a transparent liquid, what 
is essential that the solid may he entirely invisible? 

Both must have the same color and same density, or in 
other words, the same index of refraction. 



When are two points conjugate foci and when not? 

Two points, one on each side of a lens, are conjugate to 
each other when the rays diverging from one are converged to 
the other; or when one occupies the position of the object, and 
the other that of the image, and when these are interchangeable. 



The center of curvature of a concave reflectini surface is one 
meter distant; what is the power of the mirror? 

Parallel rays striking a concave surface are reflected from 
it convergently, and are made to meet at the principal focus of 
the mirror, which is just on^-half the radius. 

In this case the center of curvature or radius is one meter, 
then the principal focal distance is half-meter or twenty inches, 
which would indicate the power of the mirror to be 2 D. 



52 State Board Examinations 

When will a concave mirror not form a real focus? 

When object is placed at the principal focal distance of the 
mirror, no image is formed because the rays are reflected parallel. 

And when the object is placed closer than the principal 
focal distance, the focus is a negative one. 



Why is the power of the crystalline lens so much less when 
it is in position in the eye than it is when it is measured outside of 
the eye? 

Inasmuch as the refractive power of a convex lens is increased 
as it is moved a little farther from the eye, the power of the 
crystalline lens would be very much greater if placed in the 
position of spectacles about half-inch in front of eye than when 
it is in situ. 

Then, again, when the crystalline lens is in position in the 
eye, it is in contact with the aqueous in front, with an index 
of refraction of 1.33, and the vitreous behind with the same index: 
whereas, when it is outside of the eye it is surrounded by air, 
with an index of 1.00. Therefore there is less refraction in the 
first case when the light passes from a medium with an index of 
1.33 into one with an index of 1.43 (which is the crystalline lens), 
than in the second case above it passes from a medium with an 
index of 1.00 into one with an index of 1.43. 



What is the difference between a ray of light, a beam of light, 
and a wave of light? 

A wave of light represents the disturbance in the ether 
when light is in motion ; a ray of light is the smallest conceivable 
line of light and indicates the direction in which the light travels; 
a beam of light is a collection of parallel rays, as for instance a 
sunbeam. 



What effect do lenses have upon light coming from an object? 

Lenses bend oi refract the light and bring the rays to a 
focus and thus produce an image of the object. This applies 
only when the object is at a greater distance than the principal 



Theoretic Optics 53 

focus of the lens. If the object was at the principal focal distance, 
the rays would emerge parallel; if closer than this, they would 
emerge divergently, and in neither of these two cases would an 
image be formed. 

What is the cause of spherical aberration in lenses? 

Spherical aberration is caused by the difference in refraction 
power of different parts of the lens, as a result of which the 
rays of light passing through the lens do not all come to a focus 
at the same point, but those passing through the periphery 
come to the sooner focus. 



What is the cause of spherical aberration in mirrors? 

On account of the spherical curvature of a mirror, the rays 
of light reflected from the peripheral portions do not come to a 
focus at the same point as those reflected from the central 
portions. 

Does the speed of light increase or decrease when it passes 
from air to glass, and what is the change in the speed if the index 
of refraction of the glass is 1.50? 

The greater the density of a medium the more light is 
retarded, therefore, as it passes from air into glass its speed is 
decreased, and this decrease is to 2/3 if the index of the glass is 
1.50. 



What becomes of the energy of light when it is absorbed? 

When light is absorbed it ceases to exist as light, and it 
becomes manifest as heat. 



How can the deviating power of a 1° prism at 20 feet be cal- 
culated? 

A 1° prism will deviate light 1 cm. for each meter of distance. 
As twenty feet is equivalent to six meters, the deviation at this 
distance would be 6 cm. 



54 State Board Examinations 

What are the several meanings of the word refraction? 

Refraction means bending and has reference to the deviation 
of rays of Hght as they pass from a medium of one density into 
a medium of a different density, as for instance in passing from 
air into a glass lens and then from that lens into air, the rays are 
refracted by the lens. 

The refraction of the eye is its optical condition, or the 
manner in which it acts on the entering rays, whether bringing 
them to a focus on the retina, or in front of or behind it, these 
rays being refracted by the media of the eye. 



What relation does index of refraction hear to speed of light? 

As the index of refraction of a substance increases the speed 
of light decreases. 

How far must a convex lens of 3 inches focal distance he placed 
from a candle in order that it may produce a real image of the candle 
frame three times as large as the flame itself? 

Inasmuch as the distance of the object is to the distance of 
the image as the size of the object is to the size of the image; 
therefore, if the size of the image is to be three times the size of 
the object, its distance must be three times the distance of the 
object. 

Let X = distance of object and 3 X = distance of image, 
then 

4 X = 1/3 

X = 1/12 
3 X = 1/12 or 1/4 
Therefore the candle must be placed 4 inches from the 
lens in order that a real image may be formed, magnified three 
times. 



A parallel beam of light passes through a convex lens of power 
-\- 2 D. and which is 3 inches in diameter. What is the size of 
the circular patch of light which it casts on a screen 10 inches 
distant? 



Theoretic Optics 55 

These parallel rays of light will be brought by this lens to 
focal point at a distance of 20 inches. If these convergent rays 
are intercepted by a screen at a distance 10 inches, which is 
just one-half the distance between the lens and the focus, the 
patch of light will be exactly one-half the diameter of the lens, 
that is 1>^ inches. 



If an object is placed 12 inches in front of a plus lens whose 
focal length is 6 inches, where will the image he, and what are the 
two points called? 

Let F represent focal length and O distance of object and 
1 distance of image. Then the standard formula is 

0^1 F 

If any two of these quantities are known, the third can be 
found by this formula. In this case the distance of object and 
focal length are given, and in order to find distance of image the 
formula is 

J_ _ i_ = i. 
F 1 



or substituting figures 



_1 1_ ^ J_ 

6 12 12 



The image will be at 12 inches and the two points are called 
conjugate. , 

If the eyes of fishes focus parallel rays of light when in the 
water, what would he the refractive conditions of such eyes when 
in air? 

When the eyes of fishes are in water and light passes from 
the water into the cornea, there is no refraction because both 
have the same index of refraction, viz., 1.33. But when light 
passes from air with an index of 1.00 into the cornea with an 
index of 1.33, the rays are converged. 

Now if under the first condition parallel rays are focused on 
the retina; under the second condition they would be focused 
in front of the retina ; or in other words the refractive condition 
in air would be myopic. 



56 



State Board Examinations 



A thin convex lens having a focal 
in contact with two concave lenses 
of 7.8 inches and 12.4 inches; what 
combination? 



of 3.5 inches, is placed 
respectively focal lengths 
is the focal length of the 



We think the simplest way would be to transpose these focal 
lengths to dioptric powers and then make the necessary calcula- 
tions. 

7.8 focal length = 5.05 dioptric power 
12.4 " " = 3.17 

3.5 " " = 11.25 

Then we have + 11.25 D. placed against — 8.22 D., and 
the resultant dioptric power will be + 3.03 D., which has a focal 
length of 13 inches. 

These transpositions are made by dividing into 39.37, which 
are the number of inches in a meter. 



What is the value expressed in a single lens of the following? 

+ 2.50 cyl. axis 90° 
" = - 1.25 " " 90° 
" = - 1.00 " " 180° 
" = - 2.00 " " 180° 
" = + 1.00 " " 90° 
'' = - 0.75 " " 90° 
In order to get the proper result all the cylinders must have 
the same axis; we can transpose either to 180° or 90°. 

We will choose the latter as this requires only two transposi- 
tion. Then we have 



+ 2.50 sph. 


- 1.25 " 


+ 4.00 " 


- 2.25 " 


- 3.00 '' 


-f 1.25 '' 



-f 2.50 spl 


1. = + 2.50 cy] 


. axis 90° 


- 1.25 " 


= - 1.25 '' 


- 90° 


+ 3.00 " 


= + 1.00 '' 


- 90° 


- 4.25 '' 


= -f 2.00 '' 


- 90° 


- 3.00 " 


= + 1.00 " 


- 90° 


+ 1.25 " 


= - 0.75 '' 


- 90° 



- 1.75 sph. = + 4.50 cyl. axis 90° 
This result being obtained by algebraic addition. 



If an object at 20 inches from a convex lens has an image 20 
inches on the other side, what is the power of the lens? 



Theoretic Optics 57 

Rays reaching the convex lens from a distance of 20 inches 
would call for a convex lens of 2 D. to overcome their divergence, 
and in order that an image be formed at 20 inches on the other 
side of the lens, the rays leaving the lens must have such con- 
vergence as is produced by a convex lens of 2 D. Therefore, 
in order to accomplish both of these results the lens must have 
a power of 4 D. 

Or if O is distance of object, and 1 is distance of image, and 
F is principal focal distance, the formula is 



1+-!= 1 

0^1 F 



or substituting numbers 



20 ^ 20 20 ^^ 10 
Therefore 10 inches is the focal distance of the lens. 



Why ought a person totally immersed in water to wear convex 
lenses in order to see distinctly? 

In order that refraction may take place, the rays must pass 
from a medium of one density into a medium of a different 
density; as long as they continue in the same density there is 
no refraction. 

Under natural conditions light passes from air with an 
index of 1.00 into the cornea with an index of 1.33, and is strongly 
refracted. But when immersed in water, light passes from the 
water with an index of 1.33 into the cornea with the same index, 
and without refraction. 

Now, inasmuch as the refracting power of the cornea is 
neutralized by the water in contact with it, distinct vision would 
be impossible unless this neutralization of power is compensated 
for by a strong convex lens in front of the eye. 



How jar must a — 9 D.. sphere and a -\- 8 D. sphere he placed 
from one another in order to neutralize each other? Does it make 
any difference which lens is in front of the other? 

In order to produce neutralization these two lenses must be 
separated by the distance between their focal lengths. 



58 State Board Examinations 

Approximately the focal length of the + 8 D. lens is 5 
inches and of the — 9 D. lens, 4^ inches, and the difference 
between the two is one-half inch, and roughly speaking the 
distance between the lenses should be one-half inch. 

A more accurate way is to reduce to centimeters and subtract. 

8 D. = ^ = 12.5 cm. 

o 

9 D. = i^O = U.l cm. 

The difference between their focal lengths is 1.4 cm. 

If the convex lens is in front the parallel rays would be 
refracted so as to meet at 12.5 cm., but at a distance of 1.4 cm. 
they meet the concave lens. At this point their convergent value 
is reduced to (12.5 — 1.4 cm.) 11.1 -cm. and as the concave lens 
has an equivalent value in divergence, there is neutralization. 

If the concave lens is in front parallel rays would be made 
to diverge as if from 11.1 cm., but at a distance of 1.4 cm. they 
meet the convex lens. At this point their divergence has increased 
to (11.1 cm. -f 1.4 cm.) 12.5 cm., but as the convex lens has an 
equal amount of convergence, there will be neutralization. 



A thin plano-convex lens made of dense flint glass, having an 
index of refraction of 1.7, has a radius of curvature of 15 inches for 
its curved surface; how can you find its focal length? 



The formula is 



h(i + i^)(''-» 



F 



Or to express it in e very-day language, the focal distance 
equals the radius of curvature of the first surface plus the radius 
of the second surface, and their sum multiplied by the index of 
refraction less one. 

Substituting figures in the above formula: 



F Vl5 ^ / 



(1.7 - 1.) 



f=i3 xi^=no°^2'-^'"^^"^ 



Theoretic Optics 59 

What is the formula that applies to the conjugate focal distances 
of a spherical mirror? 

Conjugate foci are the positions occupied by the object and 
its image. They are interchangeable and when the object 
is at one, the image must be at the other. The rays diverging 
from one focus are converged to the other. 

The letter F is used to represent the principal focal distance 
of a mirror, and its reciprocal (written p) would indicate its 
reflecting power. For instance, if 8 inches was the principal 
focal distance of a mirror, its power of reflection would be Vs. 

If fi be used to represent the distance of the object, then 

its reciprocal (j) would be the power that brings parallel rays 
to a focus at such distance. 

If f2 be used to represent the distance of the image from 

the mirror, then its reciprocal (f) would indicate the power 
which causes parallel rays to meet at such distance. 

With a concave mirror the quantities are positive, and the 

full power of the mirror (represented by p) is equal to the sum 
of the powers which represent the distances of the object and 
image; or in other words, the reciprocal of the principal focal 
distance is equal to the sum of the reciprocals of any pair of 
conjugate foci, as follows: 

^ =^+ - 

F fi fa 

or 

1 1_ ^ J_ 

F fi " f. 

This is one of the most important formulae in optics. If 
any two of these quantities are known, the third can always be 
found. 

The formula is universal and holds good for all spherical 
mirrors, both concave and convex, and also for lenses. 



In what other way may the formula for conjugate focal 
distances he expressed? 

Tf the letter A be used to represent the distance of the object 



60 State Board Examinations 

from the principal focus and the letter B the distance of the 
image, then 

A B = F2 
or in other words, the distance of the object multiplied by the 
distance of the image equals the square of the principal focal 
distance. 

Suppose we have an object placed 6 inches in front of a 
concave mirror of 10 inches focal distance. 

The formula is 



1 1 1 

F f: i. 




we have 




1 1 4 
10 6 60 °^ 


1 
15 



That is the image is 15 inches behind the mirror. 

Using the second expression A B = F2 we have by substitut- 
ing figures: 

4 X 25 = 100 
and 100 being the square of 10, which is the principal focal 
distance. 



// an object he situated at sixty inches in front of a concave 
mirror of 20 inches focal length, where will the image he found? 

Here we have the principal focal distance, which is F, and 
the distance of the object which is fi. Then the reciprocal of 
the latter subtracted from the reciprocal of the former will equal 
the reciprocal of the distance of the image, as per the following 
formula: 

1 1_ ^ \^ 

F fl f2 



Substituting figures 



J__ 1 = -1 J_ 
20 60 ~ 60 °^ 30 



The image if formed at a distance of 30 inches. In this case 
30 inches and 60 inches are conjugate foci in respect to a 20- 
inch concave mirror, so that if the object is situated at a distance 
of 30 inches its image will be formed at a distance of 60 inches. 



Theoretic Optics 61 

// an object he situated at sixty inches in front of a convex 
mirror of twenty inches focal length, where will the image he found? 

Here we must keep In mind that the focal distance and 
power of a convex mirror are negative quantities, otherwise the 
rules governing the calculations of conjugate foci are the same 
as with a concave mirror. Therefore it must be expressed 



or 



The image would be a virtual one and situated 15 inches 
behind the mirror. 

— 15 inches and 60 inches are conjugate to each other with 
respect to a 20-inch convex mirror, but not 15 inches and 60 
inches; the 15 inches is a minus quantity. If the rays were 
convergent to a point 15 inches back of the mirror, on striking 
this 20-inch convex mirror their convergence would be so 
lessened that they would be reflected so as to meet at a point 
60 inches in front of the mirror. 



1 

F 


1 

fi 


+ 


1 


1 
20 


1 

60 


: — 


1 

15 



If an ohject one inch in diameter is placed 16 inches in front 
of a 24-inch mirror, what will he the size of the image? 

In the first place we must find the distance of the image, 
for which we used the regular formula: 

-1-1 =f = -1 
24 16 ' 48 

The image would be virtual and at a distance of 48 inches. 

Let O represent the size of the object, and I the size of the 
image, and fi the distance of the object and f2 the distance of 
the image. 

Then the working formula is 

- OJ2 _ 
^ fi 

Substituting figures 

1 X 48 48 



16 16 



3 inches. 



62 State Board Examinations 

The size of the image is 3 inches. 

Or we can get the result by the proportion given above, 
X : 1 ::48 : 16 

X = 3 



What is understood by Snellen's Law or the Law of Sines? 

This has reference to the determination of the index of 
refraction of a substance as compared with air, and depends upon 
the relation of the angle of incidence to the angle of refraction, 
being expressed in the following formula : 

sin i , , 

= /" (a constant) 

sin r 

This is known as Snell's Law, in honor of its discoverer, 
W. Snell, a Holland scientist, who died nearly three hundred years 
ago. 

The direction of the refracted ray is not the same as that of 
the incident ray, but there is a definite relation between these 
two directions, as shown by the law above mentioned, which 
applies to any case where a ray of light meets the dividing line 
between media of different density, as between air and glass, air 
and water, etc. 

When the light passes from a medium of less density into 
one of greater density, it is bent towards the perpendicular, when 
it follows that the angle of refraction is less than the angle of 
incidence, and under these conditions the index of refraction is 
greater than unity. 

When light passes from a medium of greater density to one 
of less density, at the surface of separation of the two media, 
it is bent away from the perpendicular, and then the angle of 
refraction is greater than the angle of incidence, under which 
conditions the index of refraction is less than unity. 

Inasmuch as nearly all transparent media are denser than 
air, their refractive indices are greater than unity, as shown by 
their deflection of light towards the perpendicular. 



What is the focal distance of a lens with an index of refraction 
of 1.54 and its two surfaces having radii of 8 inches and 5 inches? 

Let /i represent the index of refraction of the lens, r the radius 



Theoretic Optics 63 

of curvature of the first surface, and r the radius of the second 
surface; then the power of the first surface is represented by the 
formula 

u - 1 
r 

and the power of the second surface by 

P- - 1 
r' 

and the total power of the lens by 

M - 1 , M - 1 _ 1 

r "^ r' F 

or it may be written 

F = '-^ 

(r + r' ) ( /i - 1 ) 

Substituting figures 

8X5 40 . „ . , 

^ = (8+5) (1.54-1) = 7:02 = ^'^ ^"^^^^ 

Both surfaces are of the same nature and the focus is positive. 



What is the focal distance of a lens with an index of refraction 
of 1.60 and its two surfaces having radii of minus 8 inches and plus 
4 inches. 

If one surface of the lens is convex and the other surface 
concave, the focus would be real or virtual according as the con- 
vex or the concave curvature was the greater. 

We use the same formula as in the previous question: 

f = (_8+4)a.60-l) = - 4"x^.60 = - 1^-^ '"'^'■"^ 

This would mean a periscopic concave lens and the focus 
would be a negative one. 



What woidd he the curvature of a meniscus lens on its concave 
surface if the convex surface had a curvature of 5 inches, its focal 
distance was at 12 inches, and its index of refraction was 1.60? 

In order to ascertain the radius of curvature of one of the 
surfaces, when that of the other and the index of refraction and 



64 State Board Examinations 

the focal distance are known we can make use of the same formula, 
substituting the values of the known quantities and thus find the 
value of the unknown quantity, 

12 = 'S 

(5 + r) (.60) 

or 5 r = 12 (3 + .6 r) 
or 5 r = 36 + 7.2 r 
or — 2.2 r = 36 
or r = — 16.36 inches 



What is the dioptric power or number of a lens whose radii of 
curvature are 10 cm. positive and 40 cm. negative, and the index of 
refraction 1.54? 

In this case the formula is 

^ ^ 100 (m - 1) (r + r') 
r r' 

Substituting figures 

p. ^ (1 00 X .54) (10 + ( - 40) ) 
10 X - 40 

^ 54 X - 30 ^ - 1620 
- 400 - 400 

D = + 4.05 D. 



Find the dioptric number of power of a lens when the object 
is placed 50 cm. in front of it and the image 12.5 cm. behind it? 

50 cm. represents a dioptric power of 2, and 12.5 cm. a 
dioptric power of 8, and the sum of the two represents the dioptric 
power of the whole lens, viz., 

8 + 2 = 10 D. 



Find the index of refraction of a periscopic lens whose focal 
distance was 24 cm., and its radii of curvature respectively + 6 and 
— 12 cm.? 

24 = „ ^><-" .. or 24 " '^ 



(6 - 12) (" - 1) - 6 M + 6 
72 = - 144 At + 144 or - 216 = - 144 u 
or 144 M = 216 or M = 1.50. 



Theoretic Optics 65 

The index of refraction in this case is 1.50. 



A parallel beam of light passes through a convex lens of 2 D., 
ivhicli is three inches in diameter; what is the shape and size of the 
patch of light it casts on a screen ten inches away? 

This being a spherical lens the rays of light will be equally 
refracted in all meridians and will be brought to a focal point 
twenty inches from the lens. As it leaves the lens the patch of 
light will be circular in shape and the same size as the lens, that 
is three inches in diameter, and gradually lessens until it is con- 
tracted to a point at the focal distance. At a distance of ten 
inches, which is one-half the principal focal distance, it will have 
lessened one-half and form a circle of light one and a half inches 
in diameter. 

A sphero-cylindrical lens -\- 5 D. S. = — 1.50 D. C. axis 90° 
is two inches in diameter. If a parallel beam of light passes through 
this lens, what is the shape and size of the patch of light it casts on a 
screen 15 cm. away? 

In the vertical meridian the lens will have the full power of 
the sphere and will bring parallel rays of Hght to a focus at a dis- 
tance of 20 cm. 

In the horizontal meridian the power of the sphere will be 
reduced by the concave cyHnder to + 3.50 D., thus causing 
parallel rays of light to meet in a focus at a distance of about 
29 cm. 

The light passing through the vertical meridian conveying to 
the focal point 20 cm. away, meets the screen at a distance of 15 
cm., and as this is ^ the distance from the lens, the patch of light 
will be lessened in the same proportion, that is to one-half inch. 

The light passing through the horizontal meridian converg- 
ing to a point 29 cm. away, meets the screen at 15 cm., which is 
15/29 of the focal distance, and therefore the patch of light will 
be reduced to 14/29 of two inches, which is 28/29 of an inch. 

The patch of light will be oval in shape, measuring one-half 
inch vertically and 28/29 of an inch horizontally. 



// an object be held 100 cm. in front of a — 5 D. lens^ what is 
the character and location of the image formed? 



66 State Board Examinations 

This is a problem referring to conjugate foci of concave 
lenses, which refract light with that degree of divergence as if the 
rays originated from the principal focal distance of the lens. 
In the case of the — 5 D. lens in the question, the rays would 
appear to diverge from a distance of 20 cm. 

If the object is nearer than infinity, the rays from it striking 
the lens are divergent, but this divergence would be increased 
by the concave lens, and would make the light appear to originate 
from a point closer than the principal focal distance of the lens. 

If D. represents the refractive power of the lens, and the 
distance of the object be expressed in diopters as di and the 
distance of the image as d2, then the formula is 

D. = di + d2 

In the case of a concave lens it must be remembered that 
the D. and d2 are negative, and the formula would be 

D. = di + (- d2) 
or d2 = D. — di. 

Substituting the figures in the above question we have 
d2=-5-l = -6D. 

And if d2 represents a dioptric power of — 6 D., then the 
distance is 100/6 or — 16.66 cm., which means a virtual image 
16.66 cm. in front of the lens. 

If rays diverge from 100 cm. to a— 5 D. lens, after passing 
through the lens they are divergent as if they came from 16.66 
cm. 

What is the general formula for calculating the size of object 
or image, and give an illustration of it? 

The relative sizes of object and image are proportional to 
their respective distances from the center of the lens, and this 
statement is true for the virtual images of both convex and 
concave lenses. 

When the object is at twice the principal focal distance, the 
size of the image is the same; when it is farther away than 
twice the focal distance, the image is smaller; when the object 
is less than twice the focal distance, the image is larger. 

Let fi represent the distance of the object, f2 the distance of 
the image, hi the size of the object and h2 the size of the image, 
then the proportions are 

fi : f2 :: hi : h2 



Theoretic Optics 



67 



from which we get the following formula: 

hi U 



h2 = 



fl 



This formula is applicable in all cases whether the image 
be real or virtual, and whether the lens be convex or concave. 

For instance, if the size of the object be 1 inch and its 
distance 5 feet, and the image be at a distance of 5 inches, then 



h2 = 



1 X 5 
60 



_5_ 
60 



12 



The size of the image in this case is 1/12 inch. 



Draw a diagram showing the construction of a real image 
hy a convex lens. 

In the construction of the image of an object formed by a 
convex lens, there are three rays which must be considered and 
followed : 

1. The ray which passes through the optical center (O c), 
which is not refracted. 

2. The ray which is parallel to the principal axis and which 
after refraction passes through F2. 

3. The ray which passes through Fi, and which after 
refraction runs parallel to the principal axis. 

The second and third rays are the ones necessary to be 
drawn in order to locate the image of a point. 



A 






t 


\F2 




fl 


^^^^ 


sCo 


V 

c 


Bi 




1 




V 




tr 


B 


/ 




Ai 



Fig. 10 

In this diagram A B represents the object and B' A' the 
inverted image. 

Draw the principal axis from the object to the image through 
the optical center. 

Draw line from A to E parallel to the axis. This line is 
then refracted and passes through F2 extending to A'. 



68 State Board Examinations 

Draw line from A to H passing through Fi. This line after 
refraction is parallel to the axis and extends to A' . 

Draw line from A to A' passing through the optical center. 

The three lines originating from A meet at A' , and therefore 
A' is the image of A. 

In the same way B' can be constructed as the image of B, 
and then -BM' represents the position and size of the real inverted 
image of the object A B. 

If an object he placed 20 inches in front of a 6 -inch convex 
lens, and it is desired to form the image at 10 inches, what power 
of lens must he added? 

If it is desired to move the image from {2 to some other 
position, say to a more distant position expressed as X, or a 
nearer position expressed as y, there must be added to the 
original lens another lens whose power would be the difference 
between 

X ""^ IT 
or 

- — and -7 — 

y U 

For instance, if {2 be at 40 cm. and it is desired to move it 
to 50 cm., which is represented by x, then {2 = 2.50 D. and X = 
2 D. and the difference — .50 D., the necessary lens to be added 
is concave because x is more distant than {2- 

For instance, again, if {2 is situated at 40 cm. and it is desired 
to bring it up to 20 cm., which is represented by y, then {2 = 2.50 
D. and y = 5 D., and the difference between the two is + 2.50 D., 
the necessary lens to be added must be convex because y is 
closer than (2. 

In the above question, in order to find the position of ^2, 
we must subtract 1/20 from 1/6 which equals 7/60, from which 
we must subtract 1/10 (the desired distance of image being 10 
inches) and the result is 1/60. Therefore, as x is more distant than 
f2 the necessary lens is a 60-inch concave to be added to the 6-inch 
convex. 

Draw a diagram showing the construction of an image by a 
concave lens. 



Theoretic Optics 



69 



In this diagram yl ^ is an object in front of a concave lens, 
the principal focus of which is at F. 

From A we draw a Hne to H, passing through the optical 
center of the lens. 




Fig. 11 



Draw another line from A to E parallel to the axis, which 
after refraction is diverged as if it came from F. 

These two lines being divergent can meet only when produced 
backward when they come together at ^i, which is the image of ^. 

Similar lines from B meet at Bi, which is the image of B. 

Therefore, Ai Bi is the complete image of the object A B. 



How is the magnifying power of lenses expressed? 

The absolute magnification of an object is expressed by the 
ratio between the angle subtended by the image at the eye and 
the angle subtended by the object at the eye. 

If the former angle is represented by ^ / and the latter by 
A then the magnification is expressed as follows: 

A I 
A 

It may also be defined as the ratio between the size of the 
image and the size of the object on a comparison of both at the 
same distance from the eye. 

The apparent magnification is the size of the object in com- 
parison with that of the image at its point of most distinct vision. 

It is more difficult to estimate magnification in relation to 
the eye than when the image is thrown upon a screen. In the 
latter case the image and the object can both be measured, and 
the first divided by the second will show the true magnification. 

But in the case of the eye we are unable to determine 
magnification by so simple a method, for various reasons. The 
nearness of distinct vision varies in dift'erent persons; so also 
does the size of the retinal image vary with accommodation and 



70 



State Board Examinations 



shape of the eye, and also the distance of the lens from the eye 
and from the object. And then besides there is a mental condi- 
tion which affects the apparent relative size. If several persons 
were asked the size of the moon as it appears to them, they would 
probably all give a different estimate. 

A' 




Fig. 12 



If d represents the distance of most distinct vision, then 
the usual formula to express the magnifying power of the lens is 



M =^fAo4 + l 



And as 10 inches is allowed for the distance of most distinct 
vision, for the average eye, we can substitute this figure and the 
formula will read 



M = 1 + 



10 



For lenses expressed in diopters, it would be 

M = 1 +^ 



By way of illustration, if we use a 5-inch convex lens, 

10 



M 



1 + 



= 3 



Such 5-inch lens is equivalent to an 8 D. lens, with which 



M = 1 +-^= 3 



Draw a diagram showing the construction of a virtual image 
by a convex lens. 



Theoretic Optics 



71 



Draw line from A to E parallel to the axis, which line after 
refraction will pass through F2. 

Draw another line from A passing through the optical center. 

These two lines are divergent as regards each other and 
could not meet to form a real image; but by producing them 
backwards they will meet at A^, which is the virtual image of A. 

Similar lines from B w^ould locate its virtual image at Bi, 
and hence ^^ ^1 is the complete virtual image of the object A B. 



What is the effect produced by altering the position of a convex 
lens, and give several illustrations? 

8 IN On I3in 




Fig. 13 
The power of a convex lens is represented by p, which is 
the reciprocal of its principal focal distance. If the latter is 
10 inches, the power is 1/10. It is self-evident that the power 
and focal distance are fixed quantities, but at the same time 
it is true that the power of the lens, in relation to a given point, 
varies with its distance from that point; that is, when a convex 
lens is moved away from a given plane, it acts with increased 
power as regards that plane, or in other words, like a lens of 
shorter focus. 



72 State Board Examinations 

And conversely, a convex lens loses in power as regards a 
certain plane when moved closer to it, or in other words, like a 
lens of longer focus. 

Referring to diagram A in the illustrations below, it is seen 
that parallel rays of light entering a 10-inch convex lens are 
brought to a focus at a distance of 10 inches from the lens. 

If a 40-inch convex lens is placed in contact with it as in 
diagram B, the two lenses will have a principal focal distance of 

1,1 5 1 . „ . , 

and then the parallel rays will be brought to a focus at a distance 
of 8 inches. The same result will be obtained if the 10-inch lens 
is moved farther away, as in diagram C, thus proving that a 
convex lens acts with increased powers as it is moved farther 
away. 

If a 40-inch concave lens is placed in contact with the 
10-inch convex lens, the combined lenses will have a principal 
focal distance of 

1 1 3 1 .... , 
10-40=40'^^T3-'^-^-'^^^^^^^^ 

and then parallel rays will be focused at 13 inches, as shown n 
diagram D. 

The same effect is produced if the 10-inch convex lens is 
moved closer, as in diagram £, thus proving that a convex lens 
acts with diminished power as it is moved closer. 

If p represents the power of a convex lens, and d the given 

distance which it is moved, then its power would be 

1 
F - d 

Using the figures in diagrams A, B and C, 

1 1 



10-2 



The shorter distance represents a stronger lens. 

If a convex lens be moved to a greater distance than its 
principal focus, not only will its converging power be neutralized, 
but it will have the effect of a concave lens. If this 10-inch 
convex lens be moved to 13 inches from a screen, the rays will 
meet at 10 inches, that is 3 inches in front of the screen, and 



Theoretic Optics 



73 



striking the screen with this divergence will have the effect of a 
three-inch concave lens, as follows: 

\r. _ . -. — — 7-; «'. e., 3 inches negative 



What is the effect produced by altering the position of a concave 
lens, and give several illustrations? 

When a concave lens is moved farther away from a point 
back of the lens, its power is decreased and it acts as a lens with 
a longer focus. 

If a concave lens is moved closer, its power is increased, and 
it acts as a lens with a shorter focus. 



I3IN. 101N.81N. 



H 



K 



lOIN. 



-131N. 



-lOlN. 



-8 IN. 



— lOlN 



Fig. 14 



When parallel rays of Hght pass through a 10-inch concave 
lens, they will be diverged as if they came from a point 10 
inches in front of the lens, as shown in diagram F. 

If a 13-inch concave lens is made use of, the negative focus 
would be 13 inches in front of the lens, as shown in digram G. 

But the same effect can be produced by moving the 10-inch 
concave lens front 3 inches, as shown in diagram H, where the 
effect of the 10-inch lens is reduced to that of the 13-inch lens 



74 State Board Examinations 

thus proving that the power of a concave lens is decreased by 
moving it away from a given point back of the lens. 

If an 8-inch concave lens be used the negative focus will be 
at 8 inches, as shown in diagram /. 

But the same effect can be produced by moving the 10-inch 
lens back 2 inches, as shown in diagram K, where the effect of 
the 10-inch lens is increased to that of the 8-inch lens, thus 
proving that the power of a concave lens is increased by moving 
it closer to a given point back of the lens. 



What is the effect produced hy two convex lenses when they are 
separated from each other? Give an example. 

The combined power of two lenses when placed close 
together is equal to the sum of the power of each, just as simply as 
2 + 2=4. 

But if the two lenses are not in actual contact, that is, if 
they are appreciably separated, the resultant effect will not be 
the same as when they are together, which fact is easy of demon- 
stration by a drawing. 

Parallel rays of light striking the first lens, which is a 16- 
inch convex lens, will be converged so as to meet at the point A , 
which is 16 inches behind it. 




16^ 3.N. 'or=^- ^JK 



Fig. 15 

But on their way when 3 inches from the first lens, they 
meet the second lens, which is a 10-inch convex lens, and if 
they would still meet at the point A, they are now converging 
to a point 13 inches (16 — 3) behind the second lens. 

Therefore, the power of the first lens at the position of the 
second lens, is increased from 1/16 to 1/13, which means that 
the resultant effect will be the same as if a 13-inch lens was 
placed in contact with the second lens, and the combined power 
of the two lenses as separated will be 

13 + 10 = 5:6 instead of ^6 + 10 = "6 

as it would be if the lenses were close together. 



Theoretic Optics 75 

If Fi represents the front lens and F2 the back lens, and d 
the distance between them, and F the focal distance of the 
combination, then the formula for finding F would be 

Fi — d^F2 F 
Substituting the figures in the above example: 



(16-3) ' 10 5.6 

The distance of F would differ considerably if these lenses 
were reversed and the 10-inch lens faced the light, and the 16- 
inch lens was 3 inches behind it. When so arranged, according 
to the formula given above, the result would be 

10 - 3 ^ 16 4.8 



What is the effect of two concave lenses when they are separated 
from each other? Give an example. 

If the two lenses are close together their combined power 
would be equal to the sum of their individual powers, as for 
instance, if a — 2D. lens was placed against a — 3 D. lens, the 
combined power would be equal to — 5 D. But if the lenses are 
separated the resultant effect will not be the same. 




Parallel rays of light striking the first lens, which is a 20- 
inch concave lens, will be diverged as if they came from a point 
20 inches in front of the lens. 

After traveling 4 inches the rays meet the second lens, which 
is 24 inches from the negative focus of the first. Therefore, we 
have a divergence of 1/24 added to a divergence of 1/10 (which 
represents the negative power of the second lens, a 10-inch con- 
cave), which equals a total divergence of 1/7. 

Using the formula previously given: 

_j_ +^ =_i 



76 State Board Examinations 

If an object be placed 40 inches in front of a 16 -inch lens 
convex, and 4 inches behind which a 26-inch concave lens is placed, 
where will the image be formed? 

When light, falling upon a lens, is divergent instead of 
parallel, the conjugate focus must be found before we can make 
use of our formula. 

In this case, where rays diverge from a point 40 inches in 
front of the lens, which is a 16-inch convex, the conjugate focus 
would be 







1 
16 


- 


1 
40 


3 

80' 


or 


1 
27 




that is at 


27 inches. 
















Then 


we have 


















27 


1 
- 4 


+ 


(- 


1 

26 


) 


= 


1 
199 


Therefore, the image 


is 


at 199 inches 





What is the effect of a convex and a concave lens when separated? 
Give several examples. 

When a convex and a concave lens of equal power are placed 
in contact, there is neutralization and parallel rays would pass 
unrefracted, but if the lenses are separated, one or the other will 
predominate, as the case may be. 

If the distance between the lenses is less than the focal length 
of the first or convex lens, the combination will be a positive one; 
if the distance between the lenses exceeds the focal distance of 
the first lens, then the combination will be negative. 




Parallel rays of light striking the first lens, which Is an 8-inch 
convex, will be converged so as to meet in focus at a point 8 
inches away. But after having traveled 4 Inches, they meet the 
concave lens of 8 inches, and at that point they have a conver- 
gence of + }i, which is partly overcome by the 1/8 divergence of 



Theoretic Optics 11 

the concave lens, causing the rays to meet at a point 8 inches 
back of the second lens. 

Using the proper formula 



!-. + (- f) = -I 



which means a positive focus at 8 inches. 

If, instead of a convex and concave lenses being of equal 
focal length, the concave be the shorter, and then when placed in 
contact, the concave lens will predominate and no real image 
can be formed. 

If now, the convex lens be moved forward towards the 
light, it will increase in power until the point is reached where 
the separation is equal to the sum of their focal lengths, and then 
neutralization will occur and the rays will issue parallel. 

If the convex lens is moved still closer to the light, it pre- 
dominates, increasing until the separation equals the focal dis- 
tance of the convex lens, when it is at its maximum. 

In the separation of a convex and a concave lens, the result 
will be different as the light strikes first the convex or the concave. 

In the above illustration, where an 8-inch convex lens was 
4 inches in front of an 8-inch concave lens, we found that F was 
at 8 inches. 

But if the first lens was the concave one, then the result 
would be 

' + i = + 1 



- 8 - 4 ' 8 '24 

that is F would be at 24 inches. And from this we learn that F 
is closer when the convex lens is nearer the light, and farther 
away when the concave lens is nearest the light. 



// an object he placed 80 inches in front of a 14-inch convex 
lens, where must a ' 10-inch concave lens he placed so as to render 
the rays parallel? 

When the rays of light instead of being parallel are diver- 
gent when they enter the lens, as in this case, we must find the 
conjugate focus before the value of d (representing the distance 
between the lenses) can be applied to the front lens of the com- 
bination. 



78 State Board Examinations 

It will be remembered that the formula for conjugate foci 
was 

-1 _ i = _L 

F fi " f2 

Substituting the figures in this question we have 

J_ _ J_ = ^ i_ 
14 80 ~ 1120°^ 17 

The image is 17 inches back of the 14-inch convex lens. 

And in order to find where the 10-inch concave lens must be 
placed to secure parallelism, we have 

17 — 10 = 7 inches, 
it must be placed 7 inches back of the convex lens. 

When the light leaves the convex lens the rays are converg- 
ing to a point 17 inches away; but at a distance of 7 inches from 
the lens they are converging to a point 10 inches away, and at 
this position a concave lens of 10 inches would render the rays 
parallel. Or in other words, the convergence to 10 inches of the 
convex lens would be neutralized by a divergence of 10 inches of 
the concave lens. 



What must he the distance between a 10-inch convex lens and a 
4-inch concave lens, in order to produce the effect of a 40-inch 
convex lens? 

It will be remembered that the formula in such a case is 

^ (Fi - d) F. 
Fi + F2 - d 

When we come to substitute the figures in the question, we find 
we have figures for all the quantities except the distance; hence, 
we must write it as follows: 

(10 - d) - 4 



+ 40 = 



or + 40 = 



10 - 4 - d 

- 40 + 4 d 
6 - d 



For those who are not thoroughly familiar with algebra we 
would say that when we multiply + 10 by — 4 the result is — 40, 
and when we multiply — d by — 4 the result is + 4 d, which gives 
the new numerator. 



Theoretic Optics 79 

And for the denominator we add + 10 and — 4 and the result 
is + 6. It might be remarked here that when no sign is placed in 
front of a number, it is understood to be a plus. 

Now, in order to get rid of the fraction we must multiply the 
above equation by 6 — d, and the result will be 
240 - 40 d = - 40 + 4 d 

Now, we must get the d's on one side of the equation and the 
numbers on the other side, and then we have 
- 44 d = - 280 
d = 6t^ 

When a plus number is carried to the other side of the equation 
it becomes a minus, and when a minus number is carried over it 
becomes a plus. 

Now, in order to verify the result we can use the formula again 
as follows: 

(10 - 6A) - 4 . 
+ 10 - 4 - 6/x 

or ^^^X^-^) = i:^ = 40 



Therefore, the distance between the 10-inch convex lens and 
the 4-inch concave lens must be 6 tt inches in order to produce 
the effect of a forty-inch convex lens. 



A point of light is situated two meters from a -\- 2 D. spherical 
lens; at what distance will the image be formed? 

If the lens is the same number concave, where and what will 
be the image? The rays proceeding from the light at a distance of 
two meters would enter the lens with a divergence of .50 D. and, 
therefore, the lens would be reduced by that much to make the 
rays parallel. In other words, inasmuch as .50 D. of the power of 
the lens is used up to take care of the entering rays, there would 
be only 1.50 D. left to act on the emerging rays, which would bring 
the light to a focus at two-thirds of a meter on the opposite side 
of the lens from the light. 

If these rays with a divergence of .50 D. enter a concave lens, 
their divergence would be increased by the amount of divergent 
power of the lens, and in this case the rays would emerge from 
this lens with a divergence of 2.50 D., which would indicate a 



80 State Board Examinations 

negative focus of two-fifths of a meter on the same side of the 
lens as the Hght. 



A -\- 3 D. spherical lens forms an image of a light at a distance 
of 50 cm. on the opposite side of the lens from the object; at what 
distance is the light from the image? 

If the image was at a distance of 50 cm. from the lens on the 
same side as the light, what would be the distance between the 
object and the image? 

A + 3 D. lens would bring parallel rays to a focus at 33 cm., 
but if the image is formed at a distance of 50 cm. it must be that 
the rays that enter the lens are divergent instead of parallel, and 
the amount of divergence would be represented by the difference 
between 3 D. and 2 D., this latter corresponding to the diver- 
gence of the rays from the image at a distance of 50 cm. The result 
is 1 D., which means a distance of 100 cm. as the location of the 
light from the lens; or a distance of 150 cm. from object to image. 

If the image was at a distance of 50 cm. from the lens on the 
same side as the light, this would mean the virtual image of' a 
convex lens, due to the fact that the object was closer to the lens 
than its principal focal distance. 

This negative focal distance of 50 cm. indicates a divergence 
of 2 D. and if this amount remained after passing through a + 3 
D., the original divergence must have been 5 D., corresponding 
to a distance of 20 cm. as the position of the object. 

And if the distance of the object is 20 cm. and of the image 50 
cm. on the same side of the lens, the distance between object and 
image must be 30 cm. 

An image of a distant point of light is formed on a screen by 
means of a convex spherical lens. If the lens is tilted so that it is 
inclined at an angle of 45 degrees to the direction of the incident 
rays, there are two positions of the screen, as it is moved closer to 
and farther from the light, in which the image takes the form of a 
luminous straight line at right angles to each other in the two posi- 
tions. What is the explanation? 

When a spherical lens is held obliquely to incident rays of 
light, and added cylindrical effect is produced, and it acts as a 



Theoretic Optics 81 

sphero-cylindrical lens. If the spherical lens be held at its focal 
distance from the screen and parallel to it, a perfect image of a 
distant point of light is formed upon the screen; but if the lens be 
held obliquely the image is distorted and elongated as if a cylinder 
had been added to the sphere. Two bright focal lines are formed 
on the screen at right angles to each other when the lens is held 
at the proper distance for each. 

The obliquity of the lens increases the refractive power of 
both meridians, but more decidedly so in the meridian at right 
angles to the axis of rotation. Therefore, the effect produced is 
that of a slightly stronger sphere combined with a cylinder whose 
axis would correspond to the axis of rotation. In like manner 
rotation of a cylindrical lens around its axis causes increased effect 
in the meridian at right angles thereto. 

If a convex sphere be held upright and parallel to a screen, 
it will form a circular image of a point of light that is on a level 
with the axis of the lens ; but if the lens be tilted around a hori- 
zontal axis, it will no longer form a circular image, but will pro- 
duce two ill-defined lines, one vertical a little closer than the 
original focus, and another horizontal much nearer the lens; the 
latter as a result of the increase of power in the meridian of rota- 
tion. This increase of power is due to the fact that the light passes 
through a greater thickness of glass when the lens is oblique than 
when it is in its proper position. 

If a + 1 D. sphere be rotated 45°, the approximate effect 
produced would be 

+ 1.20 D. S. C + 1.20 D. cyl. 
which would show a focal line in one meridian at ^3 cm. and in 
the other meridian at 42 cm. 



An object is held 80 inches in front of two lenses, the first of 
which is a 14-inch convex and the second is a 10-inch concave. 
It is desired to focus the image 40 inches behind the concave lens; 
what must be the distance between the two lenses to accomplish 
this? 

As the object is only 80 inches in front of the convex lens, 
the rays will enter it divergently and the image will be formed 
not at the principal focal distance of the lens (which is the focus 
for parallel rays), but at some farther distance. This distance 



82 State Board Examinations 

can be found by the formula for conjugate foci, as follows: 

1 1 66 ,1 
14-80= 1120°'^^°"' 17 

Therefore, the power of the lens has been reduced to that of a 
17-inch convex lens. 

Making use of the formula for the effectivity of two lenses 
when separated, and substituting the above figures we have 

(17 - d) - 10 
17 + (- 10) - d 

For the numerator w^e multiply 17 — d by — 10, and for 
the denominator we collect and simplify. The result is 

40 = - ^^Q + ^Q^ 

7 - d 

In order to get rid of fractions we multiply by 7 — d, and 
then the equation will read 

280 - 40 d = - 170 + 10 d 
Collect the d's on one side and the numbers on the other: 
_ 50 d = - 450 
d = 9 
In order to produce the desired result the interval between 
the convex and the concave lens must be 9 inches. 



An object is placed 60 inches in front of two convex lenses, the 
first of which is a 30-inch lens and the second is a 40-inch lens; 
what must he the distance hetiveen the two lenses in order that the 
image shall he formed 10 inches hehind the second lens? 

As the rays strike and enter the first convex lens with a 
divergence of Veo, the power of the lens is practically reduced by 
that much, as follows: 

A _ J_ = J_ 

30 60 ~ 60 

We now have a 60-inch convex lens and a 40-inch convex 
lens, and by making use of the formula given in the previous 
question we have 

10 = i? rJ^ ^^ or 1000—10 d = 2400— 40 d 



60 + 40 - d 



30 d = 1400 



2400 - 40d 1 .^ ^ o 

^^ ^Q = 100 - d d = 46 2/3 



Theoretic Optics 83 

These two convex lenses must be separated by a distance of 
46% inches to accompHsh the result desired. 



What does a compound microscope consist of? Give a brief 
description of it. 

The compound microscope is used to obtain a magnified 
view of small objects close at hand. It consists of two convex 
lenses which are separated from each other, the amount of 
separation being dependent upon the length of the instrument. 

The first lens is called the objective; it should be corrected 
for aberrations, and forms a real magnified and inverted image 
of an object, which should be placed just beyond its principal 
focal distance. 

The second lens is called the eye lens, or eyepiece, or ocular; 
not quite so strong as the first, by the aid of which the image is 
viewed. 

If S. O. represents size of object, D. O. its distance from the 
microscope and D. I. the distance of the image from the objective, 
then the size of the image will be equal to 

D. I. X S. O. 
D. O. 

The magnification can be increased by a greater separation 
between the two lenses, which would increase D. I. Also by 
bringing the object closer to the microscope, which would decrease 
D. O. 

If the image formed by the objective is located at the 
principal focal distance of the eyepiece, the rays would be 
parallel, the object would seem to be at infinity and there would 
be no strain on the eye of the observer. 

The small object just beyond the focal distance of the 
objective lens, forms a real image of the object, which image 
is inverted and magnified. When the eye of the observer is 
behind the eyepiece, there is seen at the distance of most distinct 
vision a more magnified image of the object, which is still inverted 
and is now virtual instead of real as formed by the objective. 



What does an opera glass consist of ? Give a brief description 
of it. 



84 State Board Examinations 

An opera glass consists of a convex lens and a concave lens 
of higher power. When the convex lens is placed in front of the 
concave lens at a distance that is equal to the algebraic sum of 
their focal lengths, the lenses neutralize each other by their 
separation. If Fi represents the focal length of the first or convex 
lens, and F2 the focal length of the second or concave lens, then 
the amount of separation is equal to Fi plus F2, and this is the 
way the glasses should be separated for an emmetrope. 

In the case of hypermetropia where the light should enter 
the eye in convergent rays in order to focus on the retina, the 
separation of the lenses should be a little greater; while in 
myopia where the light should enter the eye in divergent rays, 
in order that vision may be clear, the distance between the lenses 
should be less. 

Although the rays of each pencil emerge parallel or divergent 
from a very distant point after refraction by the two lenses, yet 
the pencils themselves are deviated so that the object appears 
under a larger angle. 

The magnification produced by an opera glass can be 
expressed by the fraction 

F2 

If the focal length of the first convex lens was 5 inches and 
if the second concave lens was 2 inches, the magnifying power is 

!=.>. 

If the position of the lenses was reversed and the concave 
lens was placed in front, then the object viewed would appear 
diminished in size in the same ratio, which in this case would be 

_2 
5 

that is the minification would be 2^ times. 



What does a stereoscope consist of? Give a brief description 
of it. 

The stereoscope is a box or frame fitted at one end with two 
sphero-prisms bases out (which are really the transposed halves 
of one large convex lens), and at the other end there is a slide that 
carries two photographs. 



Theoretic Optics 85 

These photographs are taken from sHghtly different stand- 
points, usually about 2}4 inches apart, which correspond to 
the average pupillary distance, but which may be varied for the 
degree of stereoscopic effect desired. 

On a casual glance the photographs look alike, but they are 
really dissimilar, the right one corresponding to the image 
formed in the right eye and the left one to the image formed in 
the left eye. 

As the action of a prism is to cause an apparent displace- 
ment in the direction of its apex, the lenses of the stereoscope 
cause the apparent position of each photograph to be displaced 
inwards with the result of fusing the tw^o into one, and that with- 
out any muscular effort on the part of the eyes. The single image 
that is produced is virtual and located near the point of most 
distinct vision of the observer. 

This instrument artificially produces the appearance of 
solidity and perspective that naturally results from binocular 
vision. This effect is given to flat pictures, because each eye 
obtains a view identical with what they would receive when 
viewing the objects directly. 

The distance between the prisms and the photographs is 
equal to the focal distance of the spherical lenses, and therefore 
an emmetrope will need to use neither accommodation or con- 
vergence. If hypermetropia or myopia is present, the distance 
will have to be altered accordingly. 



What does a telescope consist of? Give a brief description! of it. 

The telescope is used to obtain an enlarged view of distant 
objects, and in its simplest form consists of two convex lenses, 
the weakest of which is called the object glass or objective, and 
is turned toward the distant object that is to be viewed; while 
the stronger is called the eyepiece or eye lens and it is placed 
immediately in front of the eye. Both of these lenses should be 
corrected for spherical and chromatic aberration. 

A real inverted image of a distant object is formed by the 
objective, and this image as seen through the eyepiece is a virtual 
one, subtending an angle greater than that of the object, and it 
is upon the ratio between these two angles that the magnification 
depends. 



86 State Board Examinations 

Since the object looked at is adistant one, the rays proceed- 
ing from it are parallel, and hence the image formed by the 
objective will be at its principal focal distance. If it is desired 
that in an emmetrope the image be seen by the eye without 
accommodation, its location must coincide with the principal 
focal distance of the eyepiece. 

In other words, the parallel rays are converged by the 
objective to a focus from which they diverge to the eyepiece, 
which parallels them, so that they can be focused in the em- 
metropic eye without accommodation. Therefore, the distance 
between the lenses must be equal to their focal lengths. 

The magnification produced by a telescope when adjusted 
to suit an emmetropic eye without accommodation is equal to 

p-^ where Fi represents the focus of the objective and F2 the 
focus of the eye lens. 

In the case of a hypermetrope the telescope would be 
adjusted so that the distance between the two lenses would be 
greater than their focal lengths, under which circumstances the 
rays would enter the eyepiece diverging from a point greater 
than its focal distance, and hence would leave the lens and enter 
the hypermetropic eye as convergent rays and thus be focused 
on its retina. 

In the case of a myope, the telescope would be adjusted so 
as to lessen the distance between the lenses, and then the rays 
would enter the eyepiece diverging from a point closer than its 
focal distance, and hence would leave the rays and as such 
would be focused on its retina. 

To obtain high magnification, the focal distance of the 
objective must be as great, and of the eyepiece as short, as 
possible; for this reason in a telescope the objective must be a 
lens of long focus and the eyepiece a lens of short focus. 

The image as seen by the eye is inverted, but with respect 
to heavenly bodies this is a matter of no importance. If so 
desired, this inversion can be overcome by means of an erecting 
eyepiece, which causes a reinversion of the image, the objection 
to which, however, is that there is a loss of light owing to the 
increase of the refracting surfaces. 



How can you determine the position and magnitude of an 
image of an object placed in front of a convex le^is? An arrow 5 



Theoretic Optics 87 

inches long is placed 8 inches away from a convex lens whose focal 
length is 3 inches. Find the position and length of the image. 

First we must find the position of the image. As the arrow- 
is at a distance of 8 inches, the rays proceeding from it would 
enter the convex lens of 3 inches focal length, with a divergence 
of 1/8. After this divergence is overcome there would remain 
in the lens a convergence power of ^, which is found as follows: 

1-1=1 = _i_ 
3 8 24 ~ 4 % 

Therefore, the position of the image would be 4 4/5 inches 
from the lens. In order to find the length of the image we have 
this proportion; the distance of the object is to the size of the 
object as the distance of the image is to the size of the image. 
Substituting figures we have 

8 :5 ::4y5 :X 
Y 5X4% _ 24 . 

Therefore, the length of the image is 3 inches. 



An incandescent gas light, with a mantle 10 centimeters 
stands at the same level as a converging lens, the power of which is 
5 D., situated 6 meters to the right of the light. Find the position 
and the size of the image of the mantle. If the light is then lifted up 
1 meter above its former position, what change will take place in 
the position of the image? 

Ordinarily, in the practice of optometry, we make our tests 
of the acuteness of vision at a distance of 6 meters, on the as- 
sumption that rays proceeding from this distance are practically 
parallel. But, as a matter of fact, these rays have a slight 
divergence. 

Six meters equal 600 centimeters, and the amount of diver- 
gence would be found by dividing 600 into 100, which would be 
0.17 D. This would reduce the power of the plus 5 D. lens to 
plus 4.83 D. 

The focal distance is found by dividing 4.83 into 100, and 
the result is 20.7 centimeters, which would be the position of the 
image. 



88 State Board Examinations 

The length of the image is found by using the proportion 
as in the previous question: 

600 cm. : 10 cm. :: 20.7 cm. : X 

10 cm. X 20.7 cm. 



600 cm. 



207 cm. - ,_ 

X = Z7^ = ^-^^^ cm. 

600 cm. 



The length of the image is 0.345 cm. 

As the Hght is elevated the image would be correspondingly 
depressed. If the light is lifted one meter, which is one-sixth the 
distance of the object, the image would be lowered in the same 
proportion, that is, one-sixth the distance of the image (1/6 of 
20.7 cm.), w^hich is 3.4 cm. 



If the refractive index from air to glass is 3/2, and that from 
air to water is 4/3, what is the ratio of the focal lengths of a glass 
lens in water and in air? 

This is equivalent to saying that when light passes from air 
into glass it is subject to an index of refraction of 1.50, and 
when it passes from air into water it is subject to an index of 
1.33, but when it passes from water into glass the conditions are 
different. 

The index of refraction from one medium to another is 
equal to the refractive index of the latter divided by that of the 
former. For all practical purposes we assume air to be the unit, 
and if the index of refraction from vacuum to glass is 1.52, or 
as we ordinarily say, the index of refraction of glass is 1.52, 
then the index of refraction from air to glass is 1.52^ 1, or 1.52. 

The relative index of refraction is the expression of the 
refractivity when light passes from one dense medium into 
another dense medium, as from water into glass or glass into 
water. According to the rule mentioned above, this relative 
index is found by dividing the index of the medium into which 
the Hght passes, by the index of the medium from which it 
proceeds. 

In this question where the light passes into glass with an 
index of 3/2 or 1.50 from water, with an index 4/3 or 1.33, the 



Theoretic Optics 89 

relative index is found by dividing 3/2 by 4/3, and the result is 
9/8 or 1.125. 

Therefore, as the refractive index from air to glass is 3/2 
or 1.50, and from water to glass is 9/8 or 1.125, the refractive 
index of the glass lens in water is reduced from 1.50 to 1.125, 
the latter being one-fourth of the former; then the focal length 
of the lens in water would be four times its length in air. 



A convex and a concave lens, each 10 inches in focal length, 
are held co-axially at a distance of 3 inches apart. Find the position 
of the image if the object is at a distance of 15 inches beyond the 
convex lens, and also the position if the lenses were reversed and the 
concave lens was first. 

If the object was at infinity the focal distance of the convex 
lens would be at 10 inches, but inasmuch as the rays enter the 
lens divergently from a point 15 inches away, the focus of the 
lens would be 

"77^ — 7^ = -77 or at 30 inches. 
10 15 30 

The formula is: p _ , "^ f^ "" F 
Substituting figures 

1 1^1 1 ^ 17 ^ 1 

30 - 3 "^ 10 27 10 270 ~ 16 

or 16 inches from the concave lens. 

There would be considerable difference in the result if the 
concave lens was in front, as follows: 

If parallel rays entered the 10-inch concave lens, its negative 
focus would be 10 inches; but, inasmuch as the rays entered the 
lens with a divergence of Vio the effective focus of the concave 
lens would be 

10-15 30 ~ 6 



Using the formula we have 



_ 1+1=1- 

V ^ 10 90 



or 90 inches from the concave lens. 



90 State Board Examinations 

What is meant by the focal length of a spherical reflecting 
surface? How far from a concave mirror of radius 3 feet would you 
place an object to give an image magnified three times? Would the 
image be real or virtual? 

Parallel rays striking a concave mirror are reflected in such 
a way as to converge to a point on the axis of the mirror, which 
is called the principal focus of the mirror. 

If the mirror be convex the rays are reflected in such a 
way as to appear to diverge from a point on the axis behind the 
mirror, which is the principal focus of the convex mirror. 

The distance from the mirror of the point to which parallel 
rays converge after reflection, or from which they seem to diverge, 
is called the focal length of the mirror. 

The focal length of a mirror, whether convex or concave, 
is equal to one-half the radius. 

In this question where the radius of the concave mirror is 
said to be 3 feet or 36 inches, the focal length would be 1>2 feet 
or 18 inches. 

The magnification is the ratio of the distance of the image 
from the mirror to the distance of the object from the mirror. 
Now, then, in order that the image may be magnified three times 
(and this magnification refers to linear dimensions and not to 
area), its distance must be three times that of the object, and in 
this question the focal distance of the object and the focal 
distance of the image must equal the focal length of the mirror, 
which is 18 inches. 

Let X = distance of image 
3 X = distance of object 

Then4X = — 
18 

72 

3 X = — or — 

72 24 

The distance of the object would be 24 inches of the image 
72 inches. 

And, inasmuch as the object is situated farther away than 
the focal distance of the mirror, the image would be real. 



A bright object 4 inches high is placed on the principal axis 
of a concave spherical mirror, at a distance of 15 inches from the 



Theoretic Optics 91 

mirror; what is the position and size of the image, the focal length 
of the mirror being 6 inches? 

The first thing is to determine the position of the image, 
as follows: j ^ 9 ^ 

"6 ~ 15 " 90 " 10 

The image is located at 10 inches. 

Now, then, the distance of the object is to the size of the 
object as the distance of the image is to its size. 

The first three factors being known, the fourth is determined 
by the following proportion: 

15 : 4 : : 10 : X 

2| 











X 
15 


_10 


40 
15 


The 


size 


of the 


image 


is 2 


2/3 


inches. 



What 'is the size and position of the image of an object 1 inch 
high placed respectively at distances of 6 inches, 9 inches, 1 foot, 
and 18 inches from a concave mirror 9 inches in radius? 

If the radius of the mirror is 9 inches, its principal focal 
distance is 4>^ inches. 

If the object is at 6 inches, the image will be at 

h~ '^ = 1^ or 18 inches 

Then, as the distance of the image (18 inches) is three times 
the distance of the object (6 inches), the size of the image will be 
three times the size of the object or 3 inches. 

If the object is at 9 inches, 

J L _ 1_ 

4>^ 9 ~ 9 

the image will be at 9 inches. 

And in this case, as the distance of the image is the same as 
the distance of the object, the size of the image will be the 
same as the size of the object, viz., 1 inch. 

If the object is at 12 inches, 

J j_ ^ J^ 

4>^ 12 71 



the image will be at 7 1/5 inches. 



92 State Board Examinations 

The size of the image is found by the following proportion: 
12 : 1 :: 7 1/5 : X 

X = '-^ = 4 
12 

The size of image is 3/5 of an inch. 
If the object is at 18 inches, 

JL _ J_ = J. = i 
4K 18 18 6 

the image will be at 6 inches. 

And as the distance of the image (6 inches) is one-third the 
distance of the object (18 inches), the size of the image will be 
one-third the size of the object, or one-third of an inch. 



What is the difference between conjugate foci that are inter- 
changeable and those that are not? Give examples of each. 

Conjugate foci are two points so related to each other that 
when the object is at the one, the image is at the other. 

If F be the principal focal distance of a mirror, fi the distance 
of the object and f2 the distance of the image, then 



F 


fl ^ f2 


1 
°^ F 


1 1 

fl ~ f. 



This is an important formula in optics; and fi and {2 are 
conjugate to each other and are interchangeable. If any two are 
know^n, the third can always be found. 

Suppose the focal length of the mirror be 8 inches, and the 
object be placed at a distance of 40 inches, then 1/8 would 
represent the reflecting power of the mirror, and 1/40 the power 
which brings parallel rays to a focus at a distance of 40 inches. 
Then 

L _ 1 = i ± 
8 40 40 °^ 10 

And f2 would be 10 inches from the mirror. 

Therefore, 40 inches and 10 inches are the distances of the 
conjugate foci of this 8-inch concave mirror. If the object be 
placed at 40 inches, the image will be formed at 10 inches, and 



Theoretic Optics 93 

if the object be placed at 10 inches, the image will be formed 
at 40 inches. 

In the above case where the conjugate foci are interchange- 
able, the object is placed at a greater distance than the focal 
length of the mirror; but if the object be placed closer than the 
focal distance of the mirror, the conjugate focus is a negative 
quantity. 

For instance, if we take this same 8-inch mirror and place 
the object 5 inches in front of it, then we have 

8 5 ~ 40 °^ 13 

That is the image (if one could be formed) would be 13 
inches behind the mirror. 

We can prove this by the formula 



Substituting figures 



Here — 13 inches (but not 13 inches) is conjugate to 5 
inches with respect to an 8-inch concave mirror, and 5 inches is 
conjugate to — 13 inches, but not to 13 inches. 



1 
fl 


-i- 


1 
F 


l+l 


f 1 > 

^ 13 ; 


) = 



What are the laws of conjugate foci as they pertain to refracting 
systems? 

1. In order that the object and image may be interchange- 
able, the rays of light must follow the same path in both direc- 
tions. 

2. If the object is moved to the right, the image moves to 
the right also; or in other words, the two foci, representing the 
object and image, always move in the same direction. (In the 
case of mirrors where the image is formed by reflection, the 
two foci move in opposite directions.) 

3. If the rays proceed from a point farther away than the 
principal focus of the lens, they will be converged and the image 
be formed at the opposite side of the lens, and will be real and 
inverted. 

4. If the rays proceed from a point closer than the principal 
focus, they will be diverged as if proceeding from an object 



94 State Board Examinations 

on the same side of the lens, and the image will be virtual and 
erect. 

When parallel rays enter a convex lens and are converged, 
we say infinity and the principal focus are conjugate. 



Does a convex lens ever produce a virtual image? If so, under 
what conditions? 

When an object is placed closer to a lens than its principal 
focal distance, the divergence of the rays is so great that the lens 
is unable to overcome it, and the rays, after passing through the 
lens, would continue divei gently. There would be no real focus, 
but a negative or virtual one, which could be found by drawing 
imaginary lines from the divergent rays backward to a point 
on the same side of the lens from which they appear to come. 

For instance, if an object was placed 10 inches from a 20- 
inch lens, the rays would enter the lens with a divergence of 
1/10, which would be partly neutralized by 1/20, the full power of 
the lens, and they would continue with a divergence of 1/20, 
or as if they came from a point 20 inches back of the lens. Under 
such conditions a convex lens produces a virtual image. 



What are the laws of reflection of light? The radius of a 
concave mirror is 6 feet, and a circular disk 1 inch in diameter is 
placed on the axis of the mirror at a distance of 2 feet from it. 
Determine the size and position of the image. 

The laws of reflection are : 

1. The angle of incidence and the angle of reflection are 
the same; that is, the angle formed by the incident ray with the 
perpendicular is equal to the angle formed by the reflected ray 
with the perpendicular. 

2. The incident and the reflected rays lie in the same plane. 
These laws are true whatever be the form of the surface, 

whether plane or curved. Spherical mirrors may be concave or 
hollow toward the light, and convex or bulging toward the light. 
In regard to a concave mirror, if the object be placed at 
its principal focal distance, the rays are reflected as parallel and 
hence no image is formed. 



Theoretic Optics 95 

When the object is located beyond the focal distance of the 
mirror, the image is real; and when object is inside of the focal 
distance, the image is virtual. 

In order to find the position of the image, we must make use 
of the formula for conjugate foci. If the radius of the mxirror 
is 6 feet, the principal focus must be 3 feet, or 36 inches. 

If F is principal focus, fi is distance of object, and U is 
distance of image, then 





1 
F 


1 
fi ' 


1 


Substituting figures 










1 
36 " 


1 
24 " 


1 
^ 72 


or 72 inches, or 6 feet. 









Inasmuch as the size of the image must be to the size of 
the object as the distance of the image is to the distance of the 
object, and as the image is at 6 feet and the object at 2 feet, or 
three times the distance, therefore, the image must be three 
times the size of object, or 3 inches. 



Practical Optics 

What is the essential difference between crown glass and flint 
glass? 

Crown glass contains lime, while flint glass contains lead. 
On account of the lead, flint glass possesses greater refractive and 
dispersive powers. It is denser and heavier than crown glass, 
but softer. 

Crown glass is harder and hence does not scratch so easily, 
but at the same time is more brittle. Crown glass also has the 
advantage of lower dispersion. The index of refraction of crown 
glass is 1.52 and of flint glass 1.62. 



Which fuses at the lower temperature, crown glass or flint 
glass? 

Technically speaking, this depends upon the index of refrac- 
tion. The index of crown glass may be raised as high as the 
lowest flint, or the index of flint glass may be lowered to the 
highest crown, when they would fuse at the same time. 

But as used in the fused bifocals, the flint fuses very much 
sooner than the crown; otherwise there could not be such a thing 
as fused bifocals. 



Mention the properties that you consider most essential in 
glass to he used for making spectacle lenses. 

Optical glass must be homogeneous, that is, of the same 
density and refractive power throughout. It should be perfectly 
transparent and free from bubbles, striae and color. 

In accordance therewith we have — 1.75 D. sph. O 4.75 D. 
cyl. axis 160°. 

c + .75 D. sph. C - 2.50 D. D. cyl. axis 145°. 

97 



98 State Board Examinations 

How may it he determined whether a lens is spheric or cylindric? 

Hold the lens at some little distance from the eye and look 
through it at a straight line on a card, or the edge of a picture 
frame or window sash. Rotate the lens or give it a circular 
motion, and if there appears any break in the line a cylinder is 
shown to be present ; the break being caused by the difference in 
refraction of different parts of the lens. If the lens is a simple 
sphere no break will be caused in the line by its rotation, because 
a sphere has the same curvature in all meridians. 

Care must be taken to look through the lens at its optical 
center, otherwise the prismatic effect of the lens will be brought 
into action, and there will be a break in the line, but in this 
case the part of the line seen through the lens will be parallel 
with the line, whereas in the case of a cylinder the line as seen 
through the lens will be oblique. This illustrates what is known 
as the "twisting" action of a cylinder. 



How can the meridian of least refraction in a sphero-cylinderic 
lens he located? 

In other words, find the location of the axis of the cylinder. 
Look through the center of the lens at a straight line or edge, 
rotate the lens and note the position in which the line is con- 
tinuous, above, below, and through the lens. Make an ink mark 
on the lens at this position, and then place on a protractor scale 
and read off the number of degrees. 

This indicates the location of one of the principal meridians, 
but it may be that of least or of greatest refraction, which are 
always at right angles to each other. The amount of motion 
in these two meridians will show at once to the experienced eye 
which one is of least refraction; or if there is any doubt about 
it, each meridian may be neutralized in turn by a spherical lens. 



Express in diopters the power of a lens whose principal focal 
distance is 150 millimeters. 

6.66 D. 



Practical Optics 

Name the various parts of a spectacle frame. 
Bridge, eye wires, end pieces, screws and temples. 



99 



Has a lenticular lens any advantage other than lightness? 

We do not know of any other advantage except lightness and 
better appearance as in the doing away with the thick edges of 
strong concave lenses. On the other hand, there may be some 
disadvantage from the restricted field. 



What is meant by a 5° prism? 

This means a prism with a refracting angle of 5' 




5 



Fig. 18 



What is meant hy the term ''base curve' as applied to toric 
lenses? 

The base curve is the curve of least strength on the toric 
surface; the three most commonly used are 3 D., 6 D. and 9 D., 
of which 6 D. is the one usually met with. These are convex in 



100 State Board Examinations 

most cases, although it is sometimes desirable to grind a concave 
base curve. 



How would you determine the optical center of a lens? The axis 
of a cylinder? 

By looking through a lens at a straight line and moving the 
lens until the line is continuous above, through and below the 
lens. At every other portion of the lens, except the optical center, 
the line will appear broken. 

By looking through a cylindrical lens at the same straight 
line and rotating the lens. The portion of the line seen through 
the lens will be "twisted" or broken by this rotation. Find the 
portion of the lens where there is no break; this will be one 
principal meridian, while the one at right angles will be the other. 
One of these will be the axis which can be determined by motion 
or by neutralization, the axis being plane and having no power. 



What is the focal length of the following, lenses placed in 
apposition: + 4 sph., + 3 sph. C^ — 2 cyl. X 75, + 3.50 sph. 
O -i- 1 cyl. X 75, - 2 sph., - 4.50 sph., - 1 cyl. X 165? 

+ 4. sph. — 2. cyl. axis 75° 

+ 3. " + 1. " " 75° 

+ 3.50 " - 1. " " 165° 

- 2. " - 1. S. 

- 4.50 " 



+ 4. sph. 
- 1. " 
+ 3. sph. 



Transpose the following: 

To -\- on — equivalent: -\- 1.75 sph. O -{-2.25 cyl. X 85 

To -\- on -\- equivalent: + 3.50 sph. O — 1-25 cyl. X 145 

To sphero-cyl.form: + .75 cyl. X 90 O - 2.50 cyl. X 180 

+ 4. D. sph. C - 2.25 D. cyl. axis 175° 
+ 2.25 D. sph. C + 1-25 D. cyl. axis 55° 
+ .75 D. sph. C - 3.25 D. cyl. axis 180° 



Practical Optics 101 

What is the prismatic power of a 5 D. lens decentered 3 mm.? 

1 D. lens decentered 10 mm. affords 1° prismatic power, 
therefore 5 D. decentered 3 mm. develops 1>^° prismatic power. 



Give the dioptric value of each surface of a wafer of a cement 
bifocal, the distance correction being — 3 sph. O + J -50 cyl. X 90, 
the addition being + 2.50 D. 

+ 3 D. and - .50 D. 



In what way does chromatic aberration differ from spherical 
aberration? 

Chromatic aberration depends upon the varying degrees 
of refrangibility of the different colors of which white light is 
composed, the violet coming to a sooner and the red to a later 
focus. 

Spherical aberration depends upon the difference in refractive 
power of the several parts of a lens, the rays passing through 
the periphery coming to a sooner focus than those passing nearer 
the center. 



A plus lens has a focus of 30 cm.; the index of refraction of 
the glass used is 1.60; one surface is co7ivex and one a radius of 
12 cm.; what is the nature and power of the other surface? 

If the radius of curvature is 12 cm, and the index of refrac- 
tion 1.60, the focal length of such a lens, if plano-convex, is 
20 cm., which makes the value of the curvature of this surface 
5 D. 

But the lens itself having a focus of 30 cm. has only a value 
of ?>.?>Z D., therefore the other surface of the lens must be concave 
1.67 D. 



What is the principal focus of a lens? 

The place or distance from lens where parallel rays are made 
to meet after passing through such lens. 



102 State Board Examinations 

How much must a 4 dioptric lens he decentered so that the 
prismatic action will he 1% prism diopters? Figure out the same 
hy means of a formula. 

AID. lens decentered 10 mm. develops 1° prismatic power; 
a 4 D. lens decentered the same distance shows 4° of prismatic 
value. If 4° prism is developed by 10 mm. decentration, then the 
desired \}i° prism will be obtained by i^t of 10; or it may be 
stated as follows: as 4° is to 10 mm. so is 1>^° to the answer. 

In both cases the result is 3.12 mm. 



With a compound lens, a plus 2 sphere on a plus 3 cylinder, 
form an image of a distant arc light. As the card is moved hack and 
forth descrihe the appearance of the image and the distance at which 
the most pronounced effects are reached. 

This compound lens would show + 2 D. power in one 
meridian and + 5 D. power in the other; the latter possessing 
the sharper curve would bring the rays to a focus at 8 inches, 
and the former at 20 inches. 

The bundle of rays on account of being more strongly re- 
fracted by one meridian w^ould assume the shape of an oval, 
the long diameter being at right angles to the stronger curve, 
until finally they unite in a focal line at 8 inches. Beyond this 
point there is a place w^here the rays assume the form of a circle, 
and they become oval in the same direction as the sharper curve 
and form a focal line at 20 inches. 



What is the dioptric power of a plano-convex lens, the radius 
of curvature being 25 cm. and the refractive index 1.50? 

To find the focal length of a convex lens the rule is to divide 

the radius of curvature by twice the index of refraction less one. 

Rad. or curv. 



(2 index — 1) 
or substituting the figures 



focal length. 



25 cm. 25 

= —T' = 25 cm. 



2 (1.50 - 1) 1 

Inasmuch as a plano-convex lens is mentioned and as the 
focal length of such a lens is equal to twice the length of the 



Practical Optics 103 

radius, we multiply the above result by two, which gives 50 cm. 
as the focal length. And as the question asks for dioptric power, 
we obtain this by dividing 50 cm. into 1 meter or 100 cm., the 
result being 2 D., which is the answer asked for. 



What is the focal length of a 7 D. lens? 

In order to find the answer to this question in inches the 
usual way is to divide 7 into 40, on the basis that 40 inches 
equal one meter, but as the exact figures are 39.37 inches to the 
meter, in order to be accurate we should use these figures as 
follows : 

7)39.37 

5.62 inches, 
which is the answer. 

In order to obtain the result in the metric system we divide 
7 into 100 cm., as follows: 

7)100 



14.28 cm. 
as the answer. 



Which point of the lens surface should be determined before 
cutting and mounting the lens? How is this point to be located and 
what is it called? 

This has reference to the optical center of the lens, and is 
located by looking through the lens at a straight line and moving 
lens until the line as seen through and beyond the lens is con- 
tinuous and unbroken. This is done for both vertical and 
horizontal meridians, and the point of intersection is the optical 
center and is the point where light passes through without 
refraction at either surface. 

It can also be located by looking through the lens at the 
vertical and horizontal edges of a card and moving lens until 
these edges are continuous through and beyond the lens, and 
then the corner of the card will indicate the optical center. 



104 State Board Examinations 

Transpose the following into various possible forms, and state 

which you prefer, giving reasons for the preference: 

+ 4.00 sph. + 125 cyl. axis ISO"" 

+ 3.25 sph. - 1.75 cyl. axis 110° 

— 2.5 sph. + 3.25 cyl. axis 70° 

The first may be transposed into anothei sphero-cylinder: 
+ 5.25 D. sph. C - 1.25 D. cyl. axis 40° 
or into a cross-cylinder: 

+ 4 D. cyl. axis 40° C + 5.25 D. cyl. axis 130° 
Some persons would prefer the original formula, because it 
is least expensive; while others would give preference to the 
second formula, because of its periscopic shape. 

The second may be transposed into another sphero-cylinder: 
H- 1.50 D. sph. C + 1.75 D. cyl. axis 20° 
or into a cross-cylinder: 

+ 1.50 D. cyl. axis 110° C + 3.25 D. cyl. axis 20° 
In this case the original formula affords a periscopic shape, 
but the second sphero-cylinder is least expensive. 

The third may be transposed into another sphero-cylinder: 
+ .75 D. sph. C - 3.25 D. cyl. axis 160° 
or into a cross-cylinder: 

+ .75 D. cyl. axis 70° C - 2.50 D. cyl. axis 160° 
In this case both of the sphero-cylinders are periscopic, but 
the second one might be preferred as being somewhat lighter. 



Why is it that when an object is viewed through a minus lens 
and the lens is moved up and down, the object appears to move in 
the same direction? 

At every part of a lens, except, exactly at its optical center, 
there is both refractive and prismatic power, the latter of which 
causes displacement of objects when viewed through the different 
portions of the lens. The displacement of an object is always 
in the direction of the apex of the prism and in a concave lens 
the apex is at the center; hence, when the center of the lens is 
moved up, an object viewed through it moves up also; when 
moved down, object moves down. Therefore, with a concave 
lens the movement is always in the same direction as the lens. 



Practical optics 105 

Describe a toric lens. 

A toric lens is one in which one of the surfaces is toroid. 
A toroid surface is one that shows two curvatures like the bowl 
of a spoon, the greater curvature being at right angles to the 
lesser. This affords the effect of a sphero-cylinder all on one 
surface. The other surface is ground in a deep meniscus form. 



What is a meniscus lens? 

Meniscus comes from a Greek word, meaning crescent, and is 
applied to lenses that are ground concave on one surface and 
convex on the other. 



Name several styles of bifocals, with a brief description of 
the same. 

Cemented, bisight, Kryptok and Ultex. 

In the cemented form the distance lens is of full size, and 
the addition necessary for reading is made by means of a thin 
wafer or segment, which is cemented on by means of Canada 
balsam to the lower portion of the large lens. 

In the bisight and Ultex forms the necessary curves for 
distance and reading are ground on a single piece of glass. 

In the Kryptok form the large lens is made of crown glass 
and the small lens of flint, and the two fused on account of the 
difference in their refractive indices; it is a matter of calculation 
to obtain the desired reading power. 



Name the parts of an eyeglass. 
Spring, guards, studs and straps. 



What is the dioptric power of a lens that focuses parallel rays 
of light at 10 cm.? 

10 D. 



How can you ascertain if a lens has been properly centered? 
By looking through the lens at a straight line or a cross on 



106 



State Board Examinations 



a large card within a few feet, or hanging on the wall across the 
room. With a convex lens, if weak, these lines may be placed at 
almost any distance, but if the convex lens is strong the distance 
must be correspondingly shortened. Another point of importance 
is that the lens must not be held too close to the eye of the 
observer — not closer than ten inches, and better if at arm's 
length. 

The position of the lens is then shifted and moved to that 
place where the lines are continuous vertically and horizontally 
through and beyond the lens; in other words, where there is no 
break in the lines. If the point of crossing of the lines agrees 
with the mechanical or geometrical center, the lens is properly 
centered; otherwise not. The optical center is indicated by the 
crossing of the lines, and it can be quickly seen if this is in the 
proper position or not. 



// your trial case contained only spherical lenses, how could 
you determine the powers of the component elements of a given 
sphero-cylindric lens? 

With a sphero-cylindrical lens there is motion in all directions 
and I would first notice if the movement is with or against. In 
this way I would determine if the lens was convex or concave, 
so as to be able to pick out a neutralizing lens of the opposite 
kind. 



I D. 



2 D. 



I D. 



2 D. 



Neutralizing Lenses 



Power of Lens 



Fig. 19 



There is one meridian in which motion is greatest and one 
meridian in which it is least, and in order to locate these two 
meridians I would look through the lens at a line or a straight 
edge and rotate it. There are two positions, and two only, 



Practical Optics 107 

where the edge is unbroken, and these indicate the location of 
the meridians of least and greatest curvature. 

I would then find a lens that would destroy the motion in 
the meridian of least curvature and another lens to neutralize 
the meridian of greatest motion. The weaker lens would be the 
sphere, and the difference between the two the cylinder with its 
axis in the direction of the weakest meridian. 

Suppose the two principal meridians were vertical and hori- 
zontal and a — 1 D. neutralized the former and a — 2D. the 
latter. 

Then the value of the lens we were neutralizing would be 
+ 1 D. sph. C + 1 D. cyl. axis 90° 



What is the characteristic difference between a plane cylindric 
lens and a sphero- cylindric lens? 

In the former one surface is piano and the other ground 
cylindrical. In the latter both surfaces are curved, one being 
ground on a spherical-shaped tool and the other on a cylinder- 
shaped instrument. 

In the sphero-cylinder all meridians have power varying in 
degree; in the piano-cylinder there is one meridian without power 
— that is, in the direction of its axis. 



How are lenses for optometric practice usually numbered? 

There are two systems by which lenses are numbered — the 
older, or inch system, w^hich shows the focal distance of the lens, 
and the newer and better system, the dioptric or metric, which 
expresses the refractive power of the lens. 



Explain the dioptric system of numbering and measuring 
lenses. 

The standard of measurement, or the unit of the dioptric 
system, is a lens of 1 D., w^hich has a focal distance of one meter. 
This lens is comparatively weak and. the stronger lenses in com- 
mon use are multiples of this unit. Weaker lenses than the unit 
are expressed in decimal fraction, one-quarter, one-half and 
three-quarters of a diopter being written .25 D., .50 D., and 
.75 D. 



108 State Board Examinations 

The power of any lens expressed in diopters is the reciprocal 
of its principal focus expressed in meters; as, for instance, a 2 D. 
lens has a focal distance of half a meter, a lens of 4 D. of one- 
quarter meter. 

The two chief objects urged against the inch system — the 
dififiiculty of combining vulgar fractions and the irregular intervals 
between the lenses — are entirely removed in the dioptric system, 
where the decimal fractions can be as easily added and sub- 
tracted as whole numbers and where there is a regular graduation 
of increase of .25 D. 

What prismatic power would he produced by decentering a 
4 D. spherical lens 5 mm.? 

The rule of decentration is as follows : For every decentration 
of 10 mm. there will be as many degrees of prismatic power as 
there are diopters of refractive power. 

Therefore, a 4 D. lens decentered 10 mm. would develop a 
prismatic value of 4°; but as the question names a decentration 
of 5 mm. the prismatic power would be one-half of the amount 
named, or 2°. 

Name, with respect to their curvatures, the different kinds of 
lens surfaces used in optometric practice. 

The surfaces may be enumerated as convex, concave, 
spherical, cylindrical and toric. 

Prisms are sometimes used in optometric practice, but the 
surfaces of prisms (if uncombined with other lenses) are plane, 
not curved. 

How is it possible to tell the principal focus of a concave lens 
without the use of neutralizing or other lenses, and without the use 
of a lens measure? 

The lens is to be held in such a position in front of a screen 
that the rays from a distant source of light as, for instance, the 
sun, will pass through it and fall upon the screen. A central 
dark spot w411 be seen upon the screen with an area of brightness 
surrounding it. 

The lens is then moved to and from the screen until a point 
is found where the area of brightness is twice the diameter of 



Practical Optics 109 

the concave lens, and then measure the distance of the lens from 
the screen, which will be the principal focal distance of the lens. 



How would you he able to tell whether a prescription for a 
prism has been filled with the amount of prism ordered; no more, 
no less? 

In the first place, w^e could tell by neutralizing the prism 
with one from the trial case, placing base over apex. For in- 
stance, if we found that the line was broken and that the part 
seen through the lens was deflected to right, we would know a 
prism was present and that its base was to the left. We take a 
prism from the test case, placing it base to the right and changing 
the prism until we found one that restored the break in the 
line and made it continuous, above, below and through the lens. 

Or w^e can make use of our knowledge of the fact that rays 
are deflected in the proportion of 1 cm. for each meter of distance. 
For this purpose we have a card showing a series of parallel lines 
which are numbered and separated by an interval of 1 cm. 

Then we hold the prismatic lens over these lines and note 
the amount of displacement and this will show the strength of 
the prism. 

Suppose we have a series of vertical lines numbered from 
left to right, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, and equidistant from each 
other. The end or O line on the right is made longer than the 
others and at its foot is placed an X . The prismatic lens is held 
with its base to the right, and the X on the O line is viewed 
through it, when it will be seen that the X has been displaced 
to the left and coincides with the line marked 3, which represents 
the strength of the prism. 

If the prism is held at a distance of 1 meter the lines should 
be 1 cm. apart; if a distance of half a meter, then ^ of a cm. 
apart, and this ratio must be always maintained or the accuracy 
of the measurement is destroyed. 



Transpose the following toric lens: Minus surface 5 D., plus 
surface 6 and + 7 .50. 

+ 1 D. sph. C + 1.50 D. cyl. 



110 State Board Examinations 

A lens measure made to measure lenses made of glass with a 
refractive index of 1.52 shows one side of a lens to he — 1.25 and 
the other to he + 2.75, hut the glass of which this particular lens 
is made has an index of refraction of 1.62; what will he the radius 
of curvature of a plano-convex lens, with an index of refraction of 
1.52, which will have the same effect on light as the first-mentioned 
lens? 

The lens measure shows this lens to be a periscopic convex 
lens of 1.50 D. when used on glass with an index of refraction of 
1.62, but as this lens measure was made to measure glass with a 
refractive index of 1.52, there would be a corresponding in- 
crease in the refractive power of the lens, as show^n by the follow- 
ing proportion: 

.52 : .62 :: 1.50 :X 
in which case X = 1.78 D., which represents the real value of 
the lens. 

Now, then, the question to be soh^ed is, what is the radius 
of curvature of a lens of this power with an index of refraction 
of 1.52? In order to obtain this w^e multiply focal distance by the 
index of refraction, less unity. The focal distance of a + 1.78 D. 
lens is approximately 22 inches. The problem is 22 x .52 = 11.44 
inches, which is the radius of curvature of the lens. 



How is it possihle to measure approximately the strength of a 
biconcave lens without either neutralizing lenses or a lens measure? 

By using the surfaces of the lens as concave mirrors and 
then make calculations on the principle that the focal length of a 
concave mirror is equal approximately to one-half the radius of 
curvature. 

For instance, if each surface of the lens focuses light by 
reflection at a distance of two inches, then the radius of curva- 
ture of each surface would be 4 inches, equivalent to a 10 D. value. 

If the index of refraction of the glass is 1.50 then the problem 
is: 

(- 1.50 - 1) (- 10 D. + - 10 D.) 
or .50 X - 20 D. = 10 D. 

which would be the approximate measure of strength of such 
biconcave lens. 



Practical optics 111 

In looking through a lejis at a small distant object, and moving 
the lens from side to side, the distant object seems to have movement; 
what is the reason for this? 

At the optical center of a lens the t^vo surfaces are parallel 
for a very minute point, and as a result the axial ray passes 
unrefracted. As the optical center is departed from the surfaces 
show a curve which results in the development of refractive and 
prismatic power, increasingly so from center to periphery. 

As the lens is moved from side to side before an eye which 
is fixed on some distant object, the visual line passes through 
different parts of the lens, each in succession showing prismatic 
power in the same direction. Now the effect of a prism is to 
cause displacement of an object looked at, and it is impossible 
to view an object through a prism without such displacement 
becoming noticeable. 

The displacement is always in the direction of the apex 
of the prism and hence in a convex lens, where the apex is at the 
periphery, the motion is opposite; while with a concave lens, where 
the apex is at the center, the motion is with. 



Under what circumstances will two strong lenses, 07ie minus 
and the other plus, each of the same curvatures, neutralize each other? 

With strong lenses it is almost impossible to get perfect 
neutralization, but the thinner the lenses the more we can 
approximate neutralization. Also if the lenses to be neutralized 
are piano-spheres, so that the two plane surfaces can be placed 
in apposition. There is difffculty in finding neutralization all 
over the lens, especially in the peripheral portions, on account of 
spherical aberration. But as the center of the lens is the portion 
patient looks through, this is the part of the lens to which most 
attention is given in neutralizing. 

Explain the effect of spherical and chromatic aberration in 
lenses for spectacles and eyeglasses. 

In the weaker numbers of glasses that are used for spectacles, 
aberration is so slight that it can practically be ignored. In the 
stronger numbers used in telescopes and microscopes, aberration 
must be and is corrected. 



112 State Board Examinations 

The effect of spherical aberration is to cause the wandering 
of rays from a single focus and thus make it more or less diffuse, 
while the effect of chromatic aberration is to cause an unequal 
refraction of colors and thus give the focus a fringe of rainbow 
colors. 



Name some important points in a spectacle lens. 

It should be perfectly clear and transparent, absolutely 
without specks, flaws or bubbles, and surfaces well polished. 
It should be of the proper size to correspond to the pupillary 
distance, correctly centered to meet the visual axes and light in 
weight. 



In how many different ways can a sphero- cylinder he made up? 

There are always two ways in which a sphero-cylinder can 
be made up, one in which the cyHnder is convex and the other 
in which it is concave, the axis of cylinder in the two ways being 
at right angles to each other. The prescription can also be made 
up in the form of a toric to give the same effect: although then 
it is not, strictly speaking, a sphero-cylinder. 



Why is the image of a point formed by a cylinder a line and 
not a point? 

The bundle of rays as they strike the lens may be represented 
by a circle. If the lens be a sphere with equal power in all me- 
ridians, the circle will be focused to a point. 

If the lens be a cylinder with no power in the meridian of 
its axis, the light will be subject to the action of the meridian at 
right angles, and as the rays are converged by this one meridian 
only they are brought to a focal line in the same meridian as the 
axis. 



// a compound lens, minus on minus, were placed before you 
with the axis of the cylinder set with an inclination somewhere 
between 90° and 180°, state how you would determine by simply 
looking through the lens at the crossbars of a window frame, that 



i 



Practical Optics 



113 



the axis zvas not hetiveeji zero and 90°. Illustrate your answer with 
a diagram. 

If a straight line be looked at through a cylindrical lens, 
either simple or compound, the line will appear straight in only 
two positions of the lens, the meridian of its axis and the meridian 
at right angles to its axis. In all other positions the line will be 
inclined to one side or the other. Hence, in looking at the 




Showing Displacement of a Straight Line by Rotation of a Cylinder 

Fig. 20 

crossbars of the window, they would appear broken if the axis 
of the cylinder was oblique, whereas if the axis w^as at 90° or 180° 
the crossbars would be continuous beyond and through the lens. 

Transpose the following into a resultant prism, and indicate 
the angle of inclination of the base-apex line in standard notation 
or by diagrams: R. E. 2%° prism base in; 1° prism base up. 



114 State Board Examinations 

We make a diagram drawing straight lines proportional in 
length to the deviating powers of the prisms, arranged according 
to the directions in which the apices of the prisms point. 

In this case we have a prism 2>^° base in, which is repre- 
sented by a horizontal line 2}i cm. long as the deviating power 



^7\ 



Fig. 21. 

of the prism towards the right. From the base of this prism we 
draw a vertical line 1 cm. long to represent the deviating power 
of the prism downward. The strength of the resultant prism 
is 2)4° as indicated by the length of the dotted line, which is 
iy2 cm., and the inclination of its base-apex line can be found 
by comparing with a protractor. 



What advantage has a periscopic lens over a double convex or 
concave lens mounted in spectacles? 

The usual form of periscopic lens is ground with what might 
be called a base curve of 1.25 D. In periscopic convex lenses 
the concave curve is 1.25 D. In periscopic concave lenses the 
convex curve is 1.25 D. 

The special advantages claimed for periscopic lenses are that 
they conform to the front convex surface of the eye, allowing 
the lenses to be brought closer to the eye and still leave room 
for the play of the lashes, and also to prevent, as far as possible, 
the cylindrical effect that is developed when the visual line passes 
obliquely through the lens. 

But it has long since been recognized that the usual peri- 
scopic lenses presented these advantages but imperfectly, and 
this led to the manufacture of the deep meniscus or toric lenses, 
with a concave curve of 6 D. or more, which are fulfilling the 
requirements for periscopic lenses and are growing rapidly in 
popular favor, notwithstanding their greater expense. 



Which surface of a + 3D. sph. Z^ -\- 1 D. cyl. should he 
placed nearer the eye? 



Practical Optics 115 

The rule Is to place the greatest concave or the least convex 
surface next to the eye and in accordance therewith the + 1 D. 
cylindrical surface should face inwards. 

Perhaps it would be better to transpose this lens into a + 4 
D. sphere O — D- cyl. and place the concave cylindrical surface 
next to the eye. 



What is the effect in strong plus lewises of having them set 
nearer to the eyes in the frames than they were in the test with the 
trial case lenses? 

A convex lens Avhen set closer to the eyes loses in effective 
power. In the ordinary lenses the change is scarcely enough to 
be considered, but in strong lenses it might be well to take it 
into account. 



What difference should there he in the position of distance and 
reading glasses? 

The optical centers of reading glasses should be a little 
closer and a little lower than for distance. The plane of the 
glasses should be vertical in distance glasses, while for reading 
they should be angled slightly forward, so that in each case the 
visual line may be at right angles to the plane of the lenses. 



How is a riding frame manipulated if one ear is higher than 
the other? 

The temple on this side should be bent upwards close to joint, 
which will result in lowering this lens and make them both of 
the same height. 



Can a prismatic effect he produced by glasses that are properly 
centered? 

If the glasses were properly centered for distance there 
would be some prismatic effect at the reading point; and if 
properly centered for reading there would be some prismatic 
effect at distance. In other words, there is bound to be some 



116 State Board Examinations 

prismatic effect when glasses are used at a distance different from 
that for which they are centered. 



What measurements must he taken in fitting spectacles to face? 

Pupillary distance, height of bridge, width of base and 
inclination of bridge. To these may be added temple width 
and temple length. 

In a case where a strong concave lens is required {say 14 D.), 
but the nose is sensitive to the weight of the glasses, what form of lenses 
can be supplied which will reduce the weight to the minimum? 

Flattened face lenses in which the thick peripheral edge is 
ground down; or lenticular, in which a small scale is cemented 
on a larger lens. 

A prescription reads -\- 1.50 for distance and + 4 D.for 
near. The distance glass is supplied in the standard periscopic 
form. What will be the power of each side of the wafer as shown 
by the lens measure? 

In a periscopic convex lens the standard concave surface is 
1.25, on which the wafer is cemented. Hence its one surface 
must be + 1.25 to correspond to the surface to which it is 
cemented, and its other surface + 1-25 to make up the 2.50 D. 
that is to be added to the distance glass. J 

In using a lens measure we cannot place its points against 1 
the cemented surface of the wafer, but we can place them against 
the surface of the distance lens, and we know^ they must both 
be of sam.e value but of opposite curves. If the wafer is large 
enough we can place the points of the lens measure against its 
other surface, and thus we are able to ascertain the curvature 
of both surfaces of the wafer 



i 



Given a plus compound, explain three ways in which the two 
meridians can be located and the power of the compound measured 
or calculated. 

The easiest way would be by the lens measure. The most 
accurate way, but involving a little more trouble, would be by 
neutralization by a similar minus compound. 

Also by looking through the lens at a line or a cross and 
locating the principal meridians, and then neutralizing each in 



Practical Optics 117 

turn by a minus sphere. Obtaining thus the power of each 
meridian, transposition is made to a sphero-cyhnder. 



How can a toric lens he made and yet not he periscopic? 

Notwithstanding a popular notion to that effect, a toric 
lens does not necessarily mean a periscopic shaped lens. The 
word toric implies that two curvatures are ground on one surface, 
as instanced in the under surface of the bowl of a spoon. The 
other surface of the lens may be flat or convex, according to the 
fancy of the prescriber, but, as a matter of fact, it has been 
found convenient to grind one surface deeply concave, but this 
is not essential to a toric lens. Of course, it is understood that 
if one surface is made concave the convexity of the other surface 
must be correspondingly increased. 

If it was desired to grind -f 1 D. sph. O + -50 D. cyl. 
axis 90° in a toric form we could have the toric surface + .50 D. 
in one meridian and + 1 D- in the other meridian, and the second 
surface + .50 D. in all meridians. This would constitute a 
toric lens, but it would not be periscopic. 



What is an achromatic lens and on ivhat principle is it made? 

The dispersive power of lenses is not always in proportion 
to their refractive power. The refraction of crown glass is in 
relation to its dispersion as 152 is to 203, while of flint glass the 
ratio is 162 to 433. This shows the high dispersive power of 
flint glass. 

These facts are made use of in the manufacture of achromatic 
lenses, which are a combination of these two different kinds of 
glass, so proportioned that the dispersion of the convex lens 
shall be neutralized by the concave lens, but not all of its refrac- 
tive power, with the result of making a refractive lens, robbed of 
its dispersive power, able to make an image without colored 
edges. In such a combination the convex lens is of crown glass 
and the concave lens of flint glass. 



What is the advantage of the meniscus form of lenses? 

With flat lenses, as the eyes turn up and down, in and out, 
the line of vision passes obliquely through the lenses, causing an 
added cylindrical effect. This is to a great extent obviated by 



118 



State Board Examinations 



the meniscus-shape lenses, whose concave surface corresponds 
to some extent to the convex surface of the eye, thus affording 
better vision as the eyes turn in different directions. Also they 
can be brought closer to the eyes and thus shut out many reflec- 
tions. 

What is the rule for the decentering of lenses? If a 5 D. sphere 
is decentered 3 mm., what will he the prismatic effect? 

For every decentration of 10 mm. there are as many degrees 
of prismatic power as there are diopters of refractive power. 
A 5 D. lens decentered 10 mm. would show 5 prism diopters, 
and decentered 3 mm. would be 3 10 of 5 = lyi prism diopters. 



The finding of a certain case is O. U. + 50 sphere, combined 
with -\- 75 D. cylinder axis 90°. These are ordered in torics on 
a -\- 6 base curve. What will the le?is measure show on the two sur- 
faces of the lens when completed and in the mountings? 




650 I 650 

Curves in sphero-C3-liuder Curves in Toric form 

Fig. 12. 



If + 6 D. is to be the base curve it is the lowest curve on 
the toric surface, which would be the vertical meridian, and 
+ 6.75 in the horizontal meridian. Then, in order to afford the 
desired power in both meridians, a — 5.50 D. curve is ground on 
the inner surface. 



Practical optics 119 

The following pair of lenses are ground on a + 7 .50 D. base 
curve: 

(a) R. E., - 2.25 D. S. = + 3.25 D. C. axis 15''; 

(b) L. E., - 1.75 D. S. = + 3.75 D. C. axis 145°. 

Give all the curves of the lenses in their principal meridians, 
indicating them on crosses. 

Write an order for riding how frames, giving all the dimensions 
for a patient with a narrow pupillary distance and a broad nose. 

Pupillary distance, 2 3/16 inches; height bridge, 1/16 
inch; inclination of bridge, 1/16 inch; in width base of bridge, 
13/16; eye. 

If you should order a -{- 1 D. sphere combined with a — 1.50 
cylinder axis 90° in the form of a toric lens, what would you find 
on checking up the glasses with a lens measure? 

Unless otherwise ordered, the toric curves are usually on 
the convex surface and the base curve is + 6 D. 

On the outer surface there would be + 6 D. curve in the 
horizontal meridian and + 7.50 D. curve in the vertical meridian. 
On the inner surface there would be — 6.50 D. in all meridians. 



If you should order a -\- 1 D. sphere combined with a — 1.50 
cylinder axis 90°, the same to be made in the toric form and on a 
base curve of — 6 D., what would you find with the lens measure? 

On the inner surface there would be a — 6 D. curve vertically 
and a — 7,50 D. horizontally. On the outer surface there would 
be + 7 D. in all meridians. 



How is the optical center of a lens found? 

By holding the lens some little distance from the eye and 
looking through it at a straight line and moving the lens until 
the line is continuous above, through and below the lens. This 
is repeated with the lens turned quarter way around, and the 
point of intersection of the two lines will indicate the location 
of the optical center. Or a cross may be used and the lens moved 



120 State Board Examinations 

until the arms are continuous without and within the lens, and 
the intersection of the arms of the cross will indicate the optical 
center. 



State the powers as shown hy a lens measure in the case of a 
toric lens -\- 8 D. sph. O + ^ cyl. axis 90°, the toric being made 
with a base curve of + 6. Will such a lens be a meniscus? 

On the outer surface there will be + 6 D, power in the 
vertical meridian and + 8 D. power in the horizontal meridian. 
On the inner surface there w411 be a + 2 D. power in all meridians. 
This is not a meniscus lens. 



What is the advantage of the metric system of numbering lenses? 

It does aw^ay with all the disadvantages of the inch system. 
The unit being comparatively weak the stronger numbers are 
multiples of the unit and can be expressed in whole numbers. 
There are no vulgar fractions to be added or subtracted, but 
instead there are decimals which are easy of manipulation. The 
interval between the lenses is regular, it expresses the dioptric 
powder of the lens instead of its focal distance, and it is a uniform 
svstem all over the w^orld. 



State the powers as shown by a lens measure in the case of a 
toric lens -\- 8 D. sph. O + ^ cyl. axis 90°, the toric lens being made 
on a base curve of — 6 D. Will such a lens be a meniscus? 

On the inner or toric surface — 6 D. power in horizontal 
meridian and — 8 D. power in vertical meridian. 

On the outer surface + 16 D. power in all meridians. 
This lens will be a meniscus. 



How many refracting surfaces has a Kryptok lens in its 
reading portion? How many refracting surfaces to the distance 
part of the lens? 

In the reading portion there are three refracting surfaces, 
the two outer surfaces, which are in contact with the air, and the 
inner surface, where the flint fits into the crown glass. 



Practical Optics 121 

The distance portion has the usual two refracting surfaces 
both in contact with air. 



On what principle is based the test for finding the optical 
center of a lens? 

At the small portion of the lens occupied by the optical 
center the surfaces are assumed to be parallel and there is no 
bending of rays of light. But as this point is departed from, 
curvature comes into evidence with refractive and prismatic 
power, increasing from the center to the periphery. 



Taking a + 1.50 D. lens from the trial case, what other lens 
must he added to it to produce a positive focus of two meters? 

A focal distance of two meters, or eighty inches, is produced 
by a + .50 D. lens; hence in order to reduce this + 1.50 D. lens 
to such power we must add — ID, 



What is the dioptric power of a lens which, being decentered 
5 mm., shows a prismatic deviation of two prism diopters? 

The rule is that for every decentration of 10 mm. there 
are as many diopters of prismatic power as there are of refractive 
power. As this question calls for a decentration of 5 mm., there 
w^ould be half as many prism diopters as there are refractive 
diopters. And as the deviating power of a prism is equal to only 
one-half of its refracting angle, there would be only one-fourth 
as much prismatic deviation as there are diopters of refractive 
power. 

Therefore, in order to produce a deviation of 2 prism diopters, 
the power of the lens must be 8 D., which, decentered 5 mm., 
would show 4 diopters of prismatic power and 2 diopters of 
deviation. 



On what particidar quality of the glasses used does the optician 
depend to secure the necessary reading addition in fused bifocals? 

On the fact that the index of refraction of the reading 
addition is so much higher than that of the distance glass, that 



122 State Board Examinations 

with the same curvature on each the former would show a much 
greater refractive power, and when properly figured out would 
afford the additional power desired for reading. 



What curvature must he given to the two surfaces of a wafer to 
produce a 2.50 D. reading addition, the distance correction being 
-\- 1 D. cylinder, the lenses to he made toric on a + 6 D. base? 

As the lens is a plane cylinder and the base curve of the 
toric surface is to be + 6 D., the inner surface must be ground 
— 6 D. in order to produce neutralization in the meridian of the 
axis of the cylinder. Therefore, the one surface of the wafer 
must be + 6 D. in order to fit close to the — 6 D. surface, and 
then in order to reduce the power of the wafer to + 2.50 D. its 
other surface must be — 3.50 D. 



Why should the optic centers of a pair of lenses for close work 
he 2 to 3 mm. nearer together than the optic centers of lenses for 
distant vision? 

On account of the greater convergence of the visual lines 
in near vision, and in order to avoid prismatic effect. 



What particular advantage has the meniscus form of lens for 
bifocal glasses? 

In order to prevent the visual lines passing obliquely through 
the lenses, as will necessarily be the case with flat lenses, as the 
eyes turn down in near vision, whereas with the meniscus form 
of lenses the visual lines are more nearly at right angles to the 
plane of the glasses, which is the condition to be desired in order to 
avoid any extra cylindrical effect. 



How can you find the principal focal length in inches and in 
millimeters of a + 6 D. lens? 

In order to find the principal focal length in inches, we divide 
the 6 D. into 40 inches, and the result is approximately 6}4 
inches. 



Practical Optics 123 

In order to find it in millimeters, we divide the 6 D. into 1,000 
millimeters, and the result is approximately 166 mm. 



What is the power of each surface of a scale used in a cement 
bifocal, where the prescription is — 1 D. sph. = + 1.25 D. cyl. axis 
180° y and + 1.75 for reading? 

The scale would be cemented on the concave surface of this 
sphero-cylinder, which, being — ID., would call for + 1 D. on 
one surface of the scale, and the other surface + .75 D., to make 
up the desired power. 

Why is it necessary to locate the optical center of a lens before 
edge grinding and mounting? 

In order to prevent any prismatic effect in the completed 
lens. 

In what two ways is the strength and character of a lens found? 

By the lens measure, which is the easiest and quickest way, 
and by neutralization, which, while more troublesome, is more 
accurate. 

How may the axis of a cylindrical lens be found? 

By the lens measure, or by finding that meridian of a lens 
in which there is no motion. 

Look through the lens at a straight line and rotate the lens 
until the line as seen through and beyond the lens forms one 
unbroken line; this will be one of the principal meridians, either 
that of greatest refraction or of no refraction at all. The latter, 
being the axis, could be determined by the absence of motion. 



Suppose we wish a toric lens to have a total dioptric power 
equal to -\- 1 ^ -{- .50 cyl. axis 120°, and the lens measure to show 
the power of one surface to be — 6 D., what will be the powers of the 
two principal meridians of the toric surface? 

This can be easily worked out diagrammatically. The first 
diagram shows the power of each of the meridians of the sphero- 



124 State Board Examinations 

cylinder, and the second diagram the complete toric lens, from 
which it is learned that the powers of the two principal me- 
ridians on the toric surface are + 7 in the 120th meridian and 
+ 7.50 in the 30th meridian. 

A toric lens, which is built up from a base curve of 6 D. on the 
toric surface. The outside surface of such lens w^ould show a -j- 6 
D. curve in the 90th meridian and a + 7 D. curve in the 180th 
meridian, while the deep concave surface towards the eye would 
be - 6 D. 





Fig. 23 



To what crossed cylinder from the trial case does the following 
correspond: 30th meridian + 3.25 D., and 120th meridian 
-h 4.25 D.? 

Inasmuch as axis is always at right angles to meridian, we 
have + 3.25 D. cyl. axis 120° C + 4.25 D. cyl. axis 30°. 



In what way does crown glass differ from flint glass as to disper- 
sion and refractive index? 

Crown glass has a lower index of refraction and shows less 
dispersion, while flint glass has a higher index of refraction and a 
much greater power of dispersion. 

This difference in dispersive power is made use of in the manu- 
facture of achromatic lenses, where the convex lens is made of 
crown glass and show^s an excess of refractive over dispersive 
power, and the concave lens is made of flint glass and shows an 
excess of dispersive over refractive power. This relation is such 
that the concave lens corrects the dispersion of the convex lens 
without neutralizing all of its refraction, and the combination 
will be an achromatic lens, which is refractive but not dispersive. 



Practical Optics 125 

In prescribing a high power concave lens, how can the weight 
of glass he reduced without decreasing the size of lens? 

By grinding down or thinning the marginal parts of the lens, 
sometimes called ''flattened face," and known to the trade as 
lenticular lenses. 



How can the axis of the cylinder in a compound he located when 
the sphere is high and the cylinder is of low power? 

By looking through a pinhole the preponderance of the 
sphere is very much reduced and then the presence of the cylinder 
soon becomes evident. This is a little practical point that is not 
as generally known as it should be. 



What is decentered lens and how can it he decentered? 

A decentered lens is one in which the optical center has been 
moved away from the geometrical center, by cutting the lens 
out of the blank to one side. The amount of decentering depends 
upon the size of the blank, and as these are not very large the 
limit of decentration is about 3 mm. 



Hew is the optical center of a lens found? 

Hold the lens some little distance from the eye and look 
through it at a straight Hne, moving the lens until the line is con- 
tinuous above, below and through the lens, and mark the same 
with ink. Turn the lens around and repeat the same procedure 
with the meridian at right angles to the first. The intersection of 
the two lines will indicate the optical center of the lens. 



What is the difference in the nature of the surface of a sphero- 
cylinder and a ioric lens of the same dioptric power? 

In a sphero-cylinder one surface is ground with a spherical 
curve and the other surface with a cylindrical curve. In a toric 
lens there are two curvatures on one surface, thus giving the effect 
of a sphero-cylinder all on one surface. 



126 State Board Examinations 

What is an achromatic lens, and does it correct all chromatic 
aberration? 

An achromatic lens is a compound lens made up of a convex 
lens of crown glass with an excess of refractive over dispersive 
power, and a concave lens of flint glass with an excess of disper- 
sive over refractive power. In this way the dispersion is over- 
come without neutralizing the refraction, and the combination 
will be a refractive but not a dispersive lens, producing a white 
spot free from colored edges. We assume it corrects the chromatic 
aberration and so it does for all practical purposes, but as all 
instruments have their limitations, it may fail to do so perfectly. 



What is the purpose of bifocal lenses, and why are they some- 
times not satisfactory? 

To hold before the eyes for immediate use the two pairs of 
glasses that are necessary for distant and near vision in a presby- 
opic ametrope. 

The chief reason why they are not satisfactory is that in look- 
ing down as in walking or stepping, the person looks through the 
lower part of the lenses which are fitted for reading at 12 to 15 
inches, whereas the ground or steps are at a distance of 65 to 
70 inches. 

What are some of the advantages and disadvantages of the 
use of Canada balsam in bifocals? 

The advantage is that it is easy to use, while the disadvan- 
tages are that it does not become perfectly hard, and hence when 
glasses are heated or vigorously rubbed the wafers are apt to slip. 

A segment for reading is attached to a distance lens whose power 
on the surface to which the segment is attached is minus 1.2 S. The 
power to be added for reading is plus 1.50; what will be the power 
of each of the surfaces of the segment? 

This is simply the attachment of a segment to the usual form 
of periscopic convex lens, which always shows on its inner surface 
a curvature of — 1.25 D. As this is the surface to which the seg- 
ment should be attached, the latter must have a corresponding 



Practical Optics 127 

convex curve, which would be + 1.25 D. on one surface. To get 
the total power of + 1.50 D. which it is desired to add for reading, 
the extra power needed over the + 1.25 D.will be + .25 D., which 
will be the power of the other surface. Hence the segment will be 
+ 1.25 D. on one surface and + .25 D. on the other. 



What effect is produced hy the decentration of a lens? 

A prismatic effect, the amount of which depends upon the 
extent of the decentration, and the curvature of the lens, in the 
proportion of 1° of prismatic power for every diopter of refraction 
power, when decentered 10 mm. 



What constitutes the axis of a cylindrical lens, and how is it 
found and numbered? 

That meridian of the lens which is piano or without curva- 
ture constitutes the axis of the cylinder. In a piano-cylindrical 
lens it coincides with that meridian in which there is no motion. 
Or it may be found by looking through the lens at a straight 
line and rotating it to that position where the line is continuous 
and unbroken, and which will indicate the position of one of 
the principal meridians, the other principal meridian will be 
exactly at right angles, one of which will be the axis as indicated 
by the absence of motion. 

How should glasses he adjusted so as to serve for both distance 
and reading? 

It is impossible to adjust glasses so that they will be accu- 
rately centered both for distance and reading, as the centers of 
reading glasses should be closer and lower than those for distance. 
The only thing that can be done is to strike a compromise, and 
when the same glasses are used for both purposes to make the 
centers slightly lower and slightly closer than would be proper 
for distance, but not so low or so close as would be desirable for 
reading alone. 

Write a prescription that shall contain a cylindric element 
for both eyes, with a difference in the spheric element of at least 



128 State Board Examinations 

2 D. and with a presbyopic addition of 1.75 for each eye The lenses 
are to he made in cemented bifocal form. State what decentration 
is necessary in order that there shall be no resulting imbalance when 
the wearer is looking through the reading portion of the lens. 

We will write supposed prescription as follows: 

R. + 1.50 D. S. o + .50 D. cyl. axis 90° 
+ 1.75 D. added 

L. + 3.50 D. S. o + .50 D. cyl. axis Q0° 
+ 1.75 added 

As the pupillary distance for reading is 4 mm. less than for 
distance, there is produced in reading vision a decentration of 
2 mm. outward for each eye, which for the 2 D. value in the 
distance lens of right eye is equivalent to .4°. 

For the left eye under the same conditions the 4 D. of 
refractive power would afford a prismatic power of .8°. 

The segment must be decentered inwards in order to neutral- 
ize the prismatic effect of the distance lens. If a + 1.75 D. lens 
decentered 10 mm. produces 1.75° prismatic power, in order to 
get .4° the problem would be 

1.75 : 10 : : .4 :X 

X = -% = 2.29 mm. 
1./5 

And as the segment must be decentered inwards 2 mm. 
anyhow, the total decentration of segment for right eye would 
be 4.29 mm. 

As the distance lens for left eye is tw4ce as strong, the amount 
of decentering of segment to overcome it would be 4.58 mm., 
which added to the customary decentering of 2 mm., would 
make a total decentering of left segment 6.58 mm. 

Which side of a toric lens is the base curve, the inner or outer? 

The base curve is the least curve on the toric surface, and 
as this is usually on the convex surface the base curve is mostly 
out. But the base curve may be on the concave surface and then 
it is in. 

// a spherical lens has a deep concave curvature on one side, 
is it a toric sphere or a meniscus? 



Practical Optics 129 

The word toric implies two curvatures on the same surface, 
and therefore strictly speaking a sphere cannot be a toric. 
But, on account of the deep concave curve that is usually ground 
on toric lenses, the term has come to be applied to all lenses 
showing such a curve even though they be spheres; therefore, a 
toric sphere and a meniscus are practically the same thing. 



Explain the construction of the spherometer and state its 
limitations. 

The spherometer, or lens measure, has three projecting metal 
pins, the two outside ones fixed and the central one projecting 
beyond the other two, movable and connected with a spring, 
which controls the pointer on the dial which indicates the 
dioptric number of the lens. When used on a lens the central 
pin is depressed and causes the pointer to revolve and indicate 
the power of the lens. 

The limitations of the instrument are that it measures only 
the curvature of the surface, and takes no account of the index 
of refraction of the glass of which the lens is made. If all glass 
was made of the same index of refraction the lens measure would 
always be correct. But, inasmuch as the index varies, it would 
be correct for only one particular index, and incorrect for lenses 
of other indices. 



// a hypermetrope is wearing a pair of -\- 4 D. sph., and these 
are by mistake decentered outward 6 mm. each, what is the amount 
of prismatic effect produced, and which direction are the bases? 

On the basis of 1° for each dioptric when decentered 10 mm., 
the effect in this case would be 2.4° for each eye bases outwards. 



The lens measure shows the inner surface of a toric lens to 
be — 4 D. sphere and the outer surface a -\- 6 D. in the 90th meridian, 
+ 8.25 D. in the 180th meridian; what is the power of the lens, and 
what is its base curve? 



The base curve would be + 6 D. and the value of the lens 
+ 2 D. S. C + 2.25 D. cyl. axis W 



\o 



130 



State Board Examinations 



What lens can he combined with a -{- 2 D. S. ^ -\- 1 D. cyl. 
axis 90° that will decrease the cylinder and increase the sphere? 

If a cylinder is placed at right angles then the power of both 
sphere and cylinder is changed; if the added cylinder be concave 
the sphere is decreased and the cylinder increased. If the added 
cylinder be convex the sphere is increased and the c^-linder 
decreased. 

Suppose, for example, we add + .25 D. cyl. axis 180° to 
the given formula, then we have + 2 D. S. O + 1 D. cyl. axis 
90° C -f .25 D. cyl. axis 180°. 

By the rules of transposition this is equivalent to + 2.25 
D. S. C + .75 D. cyl. axis 90°. 



What lens can he combined with the above that will increase 
the cylinder and decrease the sphere? 

A concave cylinder with axis at right angles. Suppose we 
add — .25 D. cyl. axis 180° to the given formula, as follows: 
4- 2 D. S. C + i D. cyl. axis 90° C - .25 D. cyl. axis 180°. 

Transposing this cross cylinder and adding the sphere, we 
have + 1.75 D. S. C + 1.25 D. cvl. axis 90°. 



If you had a case calling for + 1.25 D. S. C^ + 0.75 D. cyl. 
axis 90°, and the lenses were to he torics with minus 6 D. curve next 
to the eye, what would he the curves on the outside of the lens? 

The best way is to make a cross showing the value in each 
meridian, to which the toric transposition must be made to 
conform. 





+ 1.25 

I 1 '^^ 




- 6. 

+ 7.25 






+ 1.25 


- 6. 

+ 8. 




+ .75 


Toric Curv 


es and Values 


Sph 


+ 2.00 

ero-cylindrical Values 


+ 2. 



Practical Optics 131 

In order to maintain the powers of the sphero-cyHnder, the 
curvatures of the outer surface would be + 7.25 vertically and 
+ 8 horizontally. 

This would mean a base curve of + 7.25, but as a matter of 
fact the base curve is usually 6 D. on the toric surface, instead 
of 6 D. on the spherical surface. 



A sphero-cylindrical lens has its axis set at some unknown 
angle; how would you ascertain at what angle the meridian of 
greatest refraction is placed? At what angle is the meridian of least 
refraction? Are these two angles always to he found at right angles 
to one another? If they are, explain why; and if not, why not? 

In a sphero-cylindrical lens there are two chief meridians to 
be considered ; one of least refraction and one of greatest refrac- 
tion. In order to ascertain the location of these meridians, a 
straight vertical line should be looked at through the lens; the 
edge of a window sash or a picture frame will answer. The lens 
should be held some distance from the eye and as it is rotated the 
part of the line that is seen through the lens will break away from 
that part of the Hne seen above and below the lens, and then 
come back to its original position. 

The lens is rotated to that point where the line is continuous 
above, below and through the lens, and this will represent one of 
the chief meridians of the lens, and the other meridian will be 
exactly at right angles to it. The lens is then moved along these 
two meridians, when the meridian of greatest refraction is that in 
which the movement is the most decided, and the meridian of 
least refraction that in which the movement is least noticeable. 

The meridians of least and greatest refraction are necessarily 
at right angles to each other on account of the nature of the curves, 
the former corresponding to the axis of the cylinder where the 
power is that of the sphere alone, and the latter at right angles to 
the axis where the power is equal to that of the sphere and cylinder 
combined. 



Describe very carefully what will he seen on looking at a hori- 
zontal window har against the hright sky through a thin prism held 
with its hase upwards. 



132 State Board Examinations 

That part of the bar seen through the prism will be displaced 
downwards or in the direction of the apex of the prism. 

Prisms cause dispersion of light, but this will scarcely be 
noticeable in an experiment like this. 



A person, aged sixty, brings you the following prescription for 
distant vision: 

R. E. - 2 D. sphere C + 1.50 D. cyl. axis 90° 
L. E. - 1 D. sphere C + -75 D. cyl. axis 90° 
Can you improve on this prescription without altering the power? 
What glasses would you give him for reading at 14"? 

This prescription can be transposed into 

R. E. - .50 D. S. C - 1.50 D. cyl. axis 180° 
L. E. - .25 D. S. C - .75 D. cyl. axis 180° 
But this transposition does not present any advantages over 
the original because although they might be lighter, the periscopic 
effect would be lost. For this reason we would choose the former, 
and also because of the vertical position of the axis of the cylinders, 
which is preferable. 

At the age of 60 the amplitude of accommodation of the 
average eye is about 1 D., of w^hich the person can use one-half 
for near work. In order to read at 14 inches without accommo- 
dation, a lens of + 2.75 is called for, but as this patient can supply 
.50 D. of accommodation, the presbyopic addition would be 
theoretically + 2.25, but as a matter of fact and judging from 
experience we would say that it ought to be at least + 2.50 D. 
If we add these to the first formula we have 

R. E. + .50 D. S. C + 1.50 D. cyl. axis 90° 
L. E. -f 1.50 D. S. C + .75 D. cyl. axis 90° 
If we add to the second formula we have 

R. E. + 2.00 D. S. C - 1.50 D. cyl. axis 180° 
L. E. + 2.25 D. S. C - .75 D. cyl. axis 180° 
As the second form gives periscopic shaped lenses, they 
should be given the preference. 

How would you deterrriine the focal length of a simple hi-convex 
lens with surfaces of equal curvature? Give all the practical methods 
you know of. 



I 



Practical Optics 133 

1. By neutralization with a concave lens. 

2. By allowing rays of light from an object at least 20 feet 
away to pass through the lens and form an image on a screen, 
moving the lens closer to and farther from screen until the point 
is found where image is most distinct, and then measuring the 
distance from the lens to the screen. 

3. By measurement of the curvature of each surface by the 
lens measure. 

4. If the radius of curvature was known and the index of 
refraction of the glass from which the lens is made, we can find 
the focal distance by the following rule: divide the radius of 
curvature by twice the index of refraction less one, and the result 
will be the focal distance. 



What is the composition of flint, crown and window glass? 
What are pebble lenses? 

The composition of flint glass is as follows: 

SiHca 50 parts 

Lead 30 '' 

Potash 10 '' 

Other ingredients 10 " 

100 parts 
And crown glass: 

Silica 70 parts 

Soda 10 " 

Lime 10 " 

Other ingredients 10 " 



100 parts 
Window glass is similar in composition to crown glass, but 

not quite so good quality. 

Rock crystal or quartz is a product of Nature, and when cut 

into a slab or ground to form a lens is called a pebble. 

Pebble lenses should be clear and free from striae, specks and 

flaws and should be axis cut. 



A prescription calls for + / O + 5(9 X 135 in toric lenses, 
minus 6 base curve. What will the lens measure show on two 
surfaces? 



134 State Board Examinations 

On the concave surface towards the eye — 6 D. in the 45th 
meridian and — 6.50 D. in the 135th meridian; on the convex 
surface + 7.50 in all meridians. 



Sphero cylinders have two focal lines. What must he the 
condition so that these lines will have an actual existence? 

The sphero cyHnder must be convex and then rays of light 
passing through it will be made to meet at the focal distances of 
the two principal meridians in the form of focal lines. 



What is the character of the two focal lines in a piano cylinder? 

In the meridian of the axis, the surface of the lens is piano 
and there would be no action on light. In the meridian at right 
angles to the axis the rays would be refracted to a focal line at 
the principal focal distance of this meridian, the line being in 
the same direction as the axis. 



The following glass is called for: + 4.5 sph. O — J cyl. axis 
45°. What will the measurements be if this is made up as a toric 
on a 6 -base curve? 

This will depend upon whether it is desired to have the 
6 D. base curve on the concave or the convex surface. 

If the base curve is convex we would have + 6 D. in the 
135th meridian and + 9 D. in the 45th meridian, and the inner 
surface a — 4.50 D. curve. 

If the base curve is concave, — 6 D. in the 45th meridian 
and — 9 D. in the 135th meridian, and the outer surface a + 
10.50 D. curve. 



What is the effect produced by decentering and what is the rule 
governing the same? 

When a ray of light passes through the peripheral portion 
of a lens it is bent the same as if a prism was used, the prismatic 
effect increasing with the distance from the optical center. 



Practical Optics 135 

To produce the effect of a prism with its base in a certain 
direction, a convex lens is decentered in that same direction and 
a concave lens in the opposite direction. For example to assist 
convergence a convex lens is decentered in towards nose and a 
concave lens decentered out towards temple. 

The amount of prismatic power that may be developed by 
decentering of a lens, depends upon its strength and the amount 
of decentering, and this can be best illustrated by the prism 
diopter system of numbering. 

Since a + 1 D. lens decentered 1 cm. produces a prismatic 
power of 1 A ; 

And if P represents the prismatic power desired, D the 
dioptric power of the lens, and C the decentration in centimeters; 

then P = DxCorC=^ 

For example if a + 4 D. lens be decentered 5 cm., then 

p = 4 X 5 = 2.0 A 
Or if the effect of 1 A is desired with a + 5 D lens, the 
amount of decentration would be 

C = i cm. or 2 mm. 



What is a mobile or rotary prism? 

When two prisms are placed in apposition, with base of 
one to the apex of the other, there will be neutralization if they 
are of equal strength. If these prisms are kept in apposition and 
revolved in opposite directions, they will produce the effect of 
a single prism, gradually increasing in strength until the bases 
of the two prisms come together, when the effect will be equal 
to the combined strength of the two prisms, as was first pointed 
out by Sir John Herschel, an illustrious English astronomer, who 
died in 1871. 

The rotary prism consists of two superimposed prisms of 
15° each, mounted in a cell of the size of a trial lens. By means 
of a milled-head screw these prisms are made to revolve against 
each other. When the base of one lies over the apex of the 
other, there is neutralization, and the indicator points to zero. 
As the screw is turned and the prisms begin to rotate, prismatic 
power is developed as shown by the indicator, 2°, 4°, 6° up to 30°, 



136 State Board Examinations 

the full power being reached when the bases of the two prisms 
are in apposition. 

This rotary prism can be used in a trial frame and can be 
placed so as to present the base of the increasing prism up or 
down, in or out. 



Give all the curves of a bifocal lens, the distance lens being 
2 D. S. C^ —ID. cyl. 90° and the power of the segment being + 
2 D. What difference would there be if the index of refraction of 
the glass used was 1.50 and 1.60 respectively? 

The curvature of the two surfaces of the distance lens would 
be as shown by the formula. The segment would be attached 
to the spherical surface, and w^ould show^ a — 2 D. curve on the 
inside and a + 4 D. curve on the outside. 

When we say a — 2 D. curve or a + 4 D. curve we usually 
mean a curve that w411 produce — 2D. power or that will show 
+ 4 D. power; but as a matter of fact, in order to produce — 2D. 
power, the radius of curvature must be 10 inches; and in order 
to produce + 4 D. power, the radius of curvature must be 5 
inches, or in other words, the focus is twice the radius. 

The above figures apply only in case the index of refraction 
of the glass is 1.50. If the glass used had an index of refraction 
of 1.60, then the radius of curvature necessary to produce -f- 4 D. 
power would be 5.45 inches. 

The standard formula to find the focal length of a lens is as 
follows : 



(r + ri) (u - 1) 
Substituting figures of the above case 

- 10 r - lOr 



20 



(- 10 + r) .6 - 6. + .6r 



Clearing of fractions we have 

- 120 + 12 r = - 10 r 

- 120 = - 22 r 

r = +5.45 inches. 



Theoretic Optometry 

An anisometrope wears on the right eye a -{- 2 and on the 
left eye a — 2; his glasses are centered properly for distance; what 
is the nature of the prismatic action of the lenses when he looks at 
a nearby object? 

As convergence comes into play and he looks at a near 
object the visual lines pass in each lens slightly to the inner side 
of the optical center. 

For the right eye as he looks through a convex lens inside 
of its center, the prismatic effect produced is that of base out. 
For the left eye looking through a concave lens at its inner side, 
the effect is prism base in. As the lenses are of the same strength 
the prismatic effect produced will be equal in each eye, and one 
will neutralize the other. 



Name and describe eight subdivisions of heterophoria. Draw 
a diagram illustrative of each condition. 

Exophoria, in which there is a tendency to deviation outward, 
and the test shows the image of right eye to the left. 



L 

Fig. 24 



Esophoria, in which there is a tendency to deviation inward, 
and the test shows the image of right eye to the right. 



L R 

Fig. 25 



Right Hyperphoria, in which there is a tendency for the right 
visual line to deviate above the left. The test shows the image 
of right eye to be below the left. (See Fig. 26, page 138.) 



137 



138 



State Board Examinations 



Left Hyperphoria, in which there is a tendency for the left 
visual line to deviate above the right. The test shows the image 
of right eye to be above the left. (See Fig. 27 below.) 



R 

Fig. 26 



L 

Fig. 27 



Right Hyperexophoria, in which there is a tendency for the 
right visual line to deviate upwards and outwards. 

The test shows the image of right eye to be below and to 
the left. 



Fig. 28 



Right Hyperesophoria, signifying a tendency for the right 
visual line upward and inward. The test show^s the image of 
the right eye to be below and to the right. 



Fig. 29 



Left Hyperexophoria, a tendency of the left visual line upward 
and outward. The test shows the image of right eye above and 
to the left. 

R 



Fig. 30 



Theoretic Optometry 139 

Left Hyper esophoria, a tendency of the left visual line upward 
and inward. The test shows image of right eye to be above and 
to the right. 

R 



Fig. 31 



Explain how an eye can have normal visual acuity and he 
ametropic. 

In hypermetropia and hypermetropic astigmatism, if of 
not too high degree and in young persons, the action of the 
accommodation would neutralize the deficiency of refractive 
power or axial length, and maintain the acuteness of vision at 
the normal standard. This imposes an unnatural tax upon the 
ciliary muscle and leads to asthenopia. 



Is there any advantage in correcting an ametropia when visual 
acuity is normal? 

Yes, in order to lessen the unnatural strain upon the accom- 
modation, and relieve the headache and other symptoms of 
asthenopia which are otherwise likely to occur. 



Why is it desirable to know the far and the near point of vision ? 

The far point in emmetropia is at infinity, in hypermetropia 
beyond infinity, and in myopia at a certain finite distance ; hence 
a knowledge of the far point indicates the character of the refrac- 
tion. 

The position of the near point represents the amplitude of 
accommodation, and indirectly by comparison with the normal 
standard at the particular age shows approximately the condition 
of the refraction. In hypermetropia and presbyopia the near 
point is of importance as showing the impairment of accom- 
modation and the necessary help that must be afforded by convex 
lenses. 



140 State Board Examinations 

What is the relation, if any, of heterophoria to heterotropia? 
Explain fully. 

Heterophoria Is a tendency to a departure from the normal 
parallelism of the visual lines, while heterotropia Is an actual 
deviation. Heterophoria Is a latent strabismus, while hetero- 
tropia Is a manifest condition. 



What influence, if any, has the correcting of errors of refraction 
on muscular deficiencies? 

When esophoria occurs In connection with hypermetropla, 
and exophoria with myopia, the convex lenses In the first case 
tend to correct the esophoria, and the concave lenses In the second 
case, the exophoria. 

When, however, esophoria Is found In connection with 
myopia and exophoria with hypermetropla, which Is contrary 
to the usual relation, then the convex lenses needed to correct 
the hypermetropla would aggravate the exophoria, and the 
concave lenses In myopia would act similarly on the esophoria. 



What muscles are taxed in (a) abduction, (b) adduction, 
(c) sursumduction? Explain fully . 

Abduction Is accomplished mainly by the action of the 
external recti muscles, which Is, however, reinforced later by 
the two obliques. Hence we would say It is the external rectus 
that Is taxed. 

Adduction Is accomplished mainly by the action of the In- 
ternal recti muscles, reinforced later In the act by the superior 
and Inferior recti. We would say It Is the internal rectus that Is 
taxed. 

In sursumduction the vertical muscles are taxed, the superior 
rectus and Inferior oblique of one eye being antagonized by the 
Inferior rectus and superior oblique of the other. 



The centers of a pair of — 3.25 D. spheric lenses mounted 
in spectacles are found to be 2 mm. farther apart than the pupillary 
centers of the wearer; what prismatic correction has he been wearing? 



Theoretic Optometry 141 

The rule is that for every decentration of 10 mm. there 
will be as many degrees of prismatic power developed as there 
are diopters of refractive power in the lens. Therefore, a decen- 
tration of a 3.25 spherical lens 10 mm. would make a prismatic 
effect of 3.25. 

Now as the pupillary distance of the spectacles is 2 mm. 
farther apart than should be there is a decentration of 1 mm. 
for each eye, representing a prism of 0.325° for each lens, and as 
it is a concave lens and the decentration is outward, the base 
of the prism is in. This is less than one-third of a degree for 
each eye. 

An eyeglass mounting containing the lens correction, 

Right- 5.00 D. cyl. X ISO"" 

Left- 3.00 D. cyl. X ISO"" 
has become bent in such a manner that the wearer is looking through 
the right lens 3 mm. above the center and through the left lens 2 mm. 
below the center; what is the resulting imbalance expressed in prism 
diopters? 

A cylinder if decentered in the direction of its axis would 
show no prismatic power, but in the direction at right angles 
to its axis the same prismatic value as a sphere of similar strength. 
Therefore, in this case we have a — 5 D. sphere decentered down- 
wards 3 mm., and a — 3 D. decentered upwards 2 mm. with the 
resulting prismatic values as follows: 

O. D. \j4 A base up 

O. S. 3/5 A base down 
or a total prismatic power of a little over 2 A. 



In which position should the apex of the prism be placed to 
tax the strength of the external rectus muscle? 

The apex of a prism calls a muscle into action and taxes 
its strength ; therefore in this case the apex should be out. 



Explain the cause of separation of the corneal images of the 
keratometer targets. 

If the mires are placed in the meridian of greatest refraction 
so that the reflected images are just in contact, there will be a 



142 State Board Examinations 

separation when the instrument is turned to the meridian of 
least curvature. 



What significance may he attached to an improvejnent in 
vision that is obtained by the pin-hole disk? 

If impaired vision is due to opacities or diseased conditions, 
no improvement follows the use of the pin-hole disk or any 
other test. But if improvement in vision is afforded by the pin 
hole, the impaired vision is shown to be due to refractive error 
and imperfect focusing upon the retina and we can expect an 
equal improvement from lenses. 



For ivhat purpose is the double prism found in some trial 
cases used, and how is it used? 

It is one of the tests for muscle imbalance. 

It is placed over one eye while the patient looks at the light. 
Care must be taken that the line of separation between the two 
prisms is directly in the line of vision, in which case two lights 
will be seen by this eye, one above the other. 

When the other eye is uncovered a third light will be seen, 
the position of w^hich will show if any heterophoria is present, and 
its character. If midw^ay between the two images and in the 
same vertical line, there is no imbalance. 

If the middle image deviates to the same side esophoria is 
present; if to the opposite side, exophoria. If it moves up, 
hyperphoria of the other eye; if it moves down, hyperphoria of 
this eye. 

For the detection of insufficiencies of the oblique muscles 
the patient looks at a straight line, which is doubled by the 
prism, the two lines being parallel to each other. When the 
other eye is uncovered a third line will be seen between the 
two and should be parallel with them. If this middle line tilts 
in either direction cyclophoria is shown. 



Looking at a small object which is 4 meters distant to what 
extent will that object seem to be displaced if a prism of a power of 
4 prism diopters is placed before the eye? Explain the reasoning. 



Theoretic Optometry 143 

A prism diopter deviates a ray of light 1 cm. for each meter 
of distance. Hence the deviation at 4 meters would be 4 cm. 
and if the prism was 4 A the deviation would be 16 cm.,w^hich 
is the answer. 



What is the size of an object 5 meters away that forms on the 
retina of a hyper ope of 3 D. an image 5 mm. in diameter? 

The proportion would be as foUow^s: As 5000 mm. is to the 
size of the object, so is 14 mm. to 5 mm. To make the calculation 
the distance of the object must be multiplied by the size of the 
retinal image, and the quantity obtained divided by the distance 
of the nodal point from the retina. 

5000 mm. X 5 mm. ,„_, 



14 m.m. 






The completed proportion 


would be 




Size Ima^e Size Object 


Distance 
No ial Point 


Distance Object 


5 mm. : 1786 mm. : : 


14 mm. 


: 5000 mm. 



What direction must rays of light take in entering a hyperopia 
eye to focus on the retina? 

With the accommodation at rest the rays must be convergent 
in order to focus upon the retina of the hypermetropic eye; and 
they are made so by the action of the convex lens that is placed 
before the eye to correct its defect. 



What direction must rays of light take in entering a myopic 
eye to he focused on the retina? 

The rays of light must be divergent in order to focus upon 
the retina of the myopic eye, and they are made so by the action 
of the concave lens placed in front of the eye to correct its defect. 



A young hyperope, having normal acuteness of vision, either 
with or without + .50 D. spheric lenses, being subjected to the 
dynamic test is found to require -\- 1 D. spheric lenses to arrest 
motion of the shadow for all proximate distances up to 25 cm.; 
what lenses should he given him for reading? 



144 State Board Examinations 

In this case + .50 D. represents the manifest hypermetropia 
and + 1 D. the total hypermetropia, showing + .50 D. of latent 
defect. 

As to what lenses should be given him for reading, we would 
say theoretically + 1 D., which would be the correction of the 
total hypermetropia. But in practice this might be modified 
by the symptoms of which the patient complains and the ampli- 
tude of accommodation he possesses. Many young hyper- 
metropes with vigorous accommodation would be able to neutral- 
ize this amount of hypermetropia without any symptoms of 
asthenopia. 

When the targets of an ophthalmometer overlap at angle 45 
degrees and separate at 135 degrees what will he the axis of the + 
cylinder. 

The meridian in which the targets overlap is the meridian 
of greatest refraction, and the meridian in which they separate 
the meridian of least refraction. Therefore, 45 degrees is the 
meridian of greatest refraction, and 135 degrees the meridian of 
least refraction; the latter is the meridian that needs the assist- 
ance of the convex cylinder, and hence its axis is placed at 45 
degrees to get its refractive power at 135 degrees. 



Explain at length the reasons for prescribing distance glasses 1 
and state the exceptions that may he made when ametropia is found. 

Distance glasses are prescribed to improve vision or to relieve 
strain. 

In the first class of cases we find myopia, in which concave 
lenses are necessary to restore distant vision to normal. 

Also hypermetropia existing in a manifest form, especially _ 
in middle-aged people, where convex lenses are needed to afford a 
good distant vision. 

Also astigmatism, especially myopic, where concave cylinders 
are required for distant vision. 

In the second class of cases, there are hypermetropia and J 
hypermetropic astigmatism, where convex spheres and convex « 
cylinders may be required for constant wear to relieve strain; 



Theoretic Optometry 145 

in slight degrees occurring in young persons with vigorous 
accommodation, there may be no symptoms of asthenopia and 
the constant wearing of glasses may be deferred. 



How should a preshyope he tested for muscle imbalance at 
14 inches; if this imbalance is found, what correction, if any, would 
you give? 

The near test should be made with the reading glasses before 
the eyes. The dot and line may be used in connection with a 
vertical prism but we prefer to use the Maddox rod and a small 
point of light 14 inches away, such as can be obtained from a 
chimney with an iris diaphragm. 

Esophoria is but seldom found and calls for no correction. 

Exophoria is much more common, but is to be corrected 
only in a small percentage of cases. This may be accomplished 
by decentering the convex lenses inwards, or if more decided 
prismatic assistance is needed, by combining prism, bases in 
with the spheres, as a rule correcting not more than one-half 
the exophoria. 



In a compound lens; a -{- 2 sphere combined with a -\- 1 cyl. 
axis 90 , at what distance from the lens will the image of a distant 
point have the form of a vertical line, and at what distance will 
it have the form of a horizontal line? 

The power of this lens would be + 3 in the horizontal 
meridian and + 2 in the vertical meridian. The first focus, 
therefore, would be that of the horizontal meridian which would 
be located at 13 inches, in the form of a vertical line; and the 
second focus would be that of the vertical meridian which would 
be located at 20 inches in the shape of a horizontal line. 



What is the principle of the construction of the ophthalmometer? 

Two bright bodies called mires are reflected from the 
surface of the cornea under examination, after first being doubled 
by the prisms in the telescope. The observer looking through 



146 State Board Examinations 

the telescope sees the corneal reflections of these mires. He con- 
siders only the two central mires, which he adjusts so that they 
are just in contact in the meridian of least refraction. As the 
instrument is turned to the meridian at right angles, the mires 
will overlap, in this way showing how much one meridian exceeds 
the other in curvature. The guide lines of the mires indicate 
the location of the two principal meridians. 

The instrument shows positively the absence or presence of 
corneal astigmatism, the amount of same and the position of the 
axis of the correcting cylinder. 



How is it possible to tell whether a case is compound, simple 
or mixed astigmatism? 

In the use of the retinoscope if one meridian was found to be 
emmetropic and the other ametropic the case is simple astigma- 
tism. If both meridians are hypermetropic or myopic, varying 
in degree, compound astigmatism. If one meridian is hyper- 
metropic and the other myopic, mixed astigmatism. 

Or by the trial case examination, if a piano cylinder was 
required, the case is one of simple astigmatism. If the sphere 
is called for in addition to the cylinder, the case is compound or 
mixed, according to whether the signs are alike or unlike. 



If a prism is placed in front of the eye, base downward, what 
will be the effect on the eye behind that prism? 

As the eye turns toward the apex of a prism in this case the 
eye will turn upwards. 

What is an ophthalmometer? Describe the principles upon 
which it is constructed and state some of the reasons why its readings 
cannot in all cases be relied upon. 

The ophthalmometer is an instrument devised to measure 
the corneal curves and consists essentially of a telescope and a 
set of mires. The latter are adjusted for the meridian of least 
refraction and then the instrument is turned 90° to the meridian 
of greatest refraction and the difference between the curvatures 
of the two meridians is shown at once. 



Theoretic Optometry 147 

As the scope of the instrument is limited to the measure- 
ment of the curvature of the anterior surface of the cornea, it is 
evident that it cannot be rehed upon in all cases of astigmatism, 
because in a limited number of cases there is a certain amount 
of lenticular astigmatism which must be taken into account in 
estimating the astigmatism of the eye as a whole. This lenticular 
astigmatism may be static or dynamic, either of which changes 
the error as indicated by the ophthalmometer. This explains 
why the eye will not always accept the cylinder representing the 
corneal astigmatism as measured by the ophthalmometer. 



What is an ophthalmometer? 

An instrument w^hich by means of reflected images measures 
the curvature of the cornea in its several meridians, and is used 
in the detection and measurement of astigmatism. It makes an 
objective test. 

What is a perimeter? 

An instrument to measure the extent of the field of vision 
and to detect the existence of scotomata. It makes a subjective 
test, but as it has reference only to indirect vision it is of no 
special value in the fitting of glasses. 



What is a phorometer? 

Consists of two prisms, one before each eye. which rotate 
in conjunction, and it can be used not only to detect any muscle 
imbalance, but to measure it at the same time. 



If a person ivore a 3 prism, base out over one eye and a 2° 
prism, base in, over the other, what would he the effect of the same? 

The effect of a prism base in is to neutralize the effect of a 
prism base out, and hence in this case the prism base in would 
neutralize 2° of the prism base out, and the result would be a 
prism of 1 base out. 



148 State Board Examinations 

If the far point with + 4.5 sphere is 60 cm., and with a + 2.5 
sphere it is at 30 cm., what is the refractive condition? 

A far point of 60 cm. would represent a myopia of 1.66 D., 
and if this condition is artificially produced by a + 4.50 D. lens 
then we must assume the eye is hypermetropic to the extent of 
the difference between them, viz., 2.84 D. 

A far point of 30 cm. would indicate a myopia of ?>.?>?> D., 
and if such myopia is artificially produced by a + 2.50 lens, 
then we infer that the refraction of the eye must be myopic to 
the extent of the difference between them, or .83 D. 



A hoy, 10 years, has a visual acuity of 6/60; with a + 3 sph. 
he can see 20/20, and the near point is at 33 cm. What is probably 
the refractive error in this case? 

In a case like this, where an acuteness of vision of only one- 
tenth is raised to normal by a convex lens, the refractive error 
is undoubtedly hypermetropia. But whether this + 3 D. 
sphere represents the full correction is another question, on 
which we have no information. If normal vision is reached with 
this lens the optometrist should not stop here, but continue 
with stronger lenses as long as a vision of 20/20 is maintained, 
because the measure of the error is the strongest convex sphere 
the patient can be induced to accept for distance. 

Taking up the question of the optics of such an eye from a 
theoretic standpoint, if the + 3 D. lenses produced a near 
point of ?)?) cm. they would seem to fall far short of a full correc- 
tion. This near point indicates an accommodation of 3 D., 
whereas the normal amplitude at this age is 14 D., showing a 
deficiency of 11 D., which added to the 3 D. would make 14 D. 
as the measure of the defect in this eye. 

Looking at the matter from a practical standpoint, in the 
presence of such a high degree of hypermetropia, it would scarcely 
be possible to raise the vision to normal by a + 3 D. lens, or, 
for that matter, by any lens. 



A patient assisted by 1 D. lenses reaches the punctum proximum 
at 33 cm.; what is his amplitude of accommodation? 



Theoretic Optometry 149 

This punctum proximum represents an amplitude of accom- 
modation of 3 D. If this was reached by the assistance of 1 D. 
lenses, then the patient's natural amplitude of accommodation 
is 2 D. 

In the above case, i^hat proportion of the accommodation is 
held in reserve luhen + 1.5 D. lenses are used for reading at 33 cm.? 

In order for an emmetrope to read at 33 cm.. 3 D. of accom- 
modation is necessary, of which amount 1.50 D. is supplied by 
the convex lenses, and the other 1.50 D. by the accommodation 
itself. As this patient is said to ha\-e an amplitude of accom- 
modation of 2 D.. the amount held in reserve is .50 D. 



A certain person requires for distance a — 1.50 sph., and has 
an amplitude of accommodation of 6 D. WitJi what glass will he 
see most clearly at a distance of 20 inches, the — 1.50 or a + .50? 

If a + .50 D. lens is placed in front of this person, who is 
myopic to the extent of 1.50. his myopia is increased to 2 D., 
and hence this specified distance of 20 inches would represent 
his far point where he would be able to see without effort of 
accommodation. If, on the other hand, the — 1.50 D. lenses 
were used, his amplitude of accommodation vrould be reduced 
to 4.50 D., of which he would require 2 D. to see at 20 inches, 
leaving 2.50 D. in reserve. 

Whether he would be able to see most clearly at 20 inches 
without any effort of accommodation, as in the one case, or at 
the expense of less than half his available accommodation, as in 
the other case, is a point that cannot be determined by any rule. 



What is the function of the retina? 

To receive the images of external objects as they are focused 
by the refracting media, and transfer the impressions it thus 
receives through the optic nerve to the brain. 



// patient's vision is 20 20 and he can see all the lines on the 
clock dial distinctly and equally, and can also read the very smallest 



150 State Board Examinations 

print at his ordinary reading distance, what refractive error cafi 
he have? 

He might have a small amount of hypermetropia, which in 
a young person is readily overcome by an active accommodation. 
It is also likely that he might have a slight hypermetropic 
astigmatism, but not possible to have myopia or myopic astigma- 
tism. 



At what distance from the patient must the test card he placed 
so that light coming from it will he practically parallel? 

By universal consent it has been agreed that the test card 
shall be placed at 20 feet and that rays therefrom shall be assumed 
to be parallel for our purposes in testing. 

Strictly speaking, rays from 20 feet have a divergence of 
240 inches, which is equal to one-sixth of a diopter, and hence 
lenses fitted at this distance might be that much too strong in 
convex and too weak in concave. 



When does the stenopaic slit act best? 

We would say in myopic astigmatism of marked degree, or 
in hypermetropic astigmatism when a cycloplegic has been 
used, as otherwise the accommodation would make its results 
uncertain. 



What is effect of pupillary distance of lenses too wide or too 
narrow? 

When too wide the effect of a prism base out in convex lenses, 
and base in with concave lenses. 

When too narrow the effect of prism base in with convex 
lenses, and base out in concave lenses. 



When should the pin-hole test he used? 

In all cases where the visual acuity is much impaired. It 
would be of no value in cases where the acuteness of vision is 
normal, or nearly so. 



Theoretic Optometry 151 

With the vertical lines in the fan the blackest, the others being 
dull, blurred or gray, what is the axis of the correcting cylinder, and 
what is the abnormal meridian? 

If the vertical lines are clearest, it is the vertical meridian 
of the eye that is abnormal, and as the axis of the cylinder is 
placed in the same direction as the indistinct lines, in this case 
the axis would be horizontal, and in this position would correct 
the abnormal vertical meridian. 



What is the purpose of the test with the Maddox rod? What 
is tested when the rod is vertical? 

The purpose of the Maddox rod is to form in the eye a 
retinal image that is so dissimilar in size, shape and appearance 
from the other that the natural desire for fusion is suppressed, 
thus giving the eyes over to the action of their muscles and 
allowing any imbalance to become manifest. 

When the rod is vertical the test is made for hyperphoria. 



State how you wotdd measure the amplitude of accommodation 
and the power of convergence? 

By measuring the closest possible point of vision with the 
smallest type and transposing the same into diopters will show 
the amplitude of accommodation. 

For instance, if 10 inches was the nearest possible reading 
point, the amplitude of accommodation would be 4 D. 

The power of convergence is measured by the strongest 
prisms bases out which the eyes are able to overcome and main- 
tain single vision of a point of light. 



In case of ametropia, by what means may a prognosis of an 
improvement in vision be made without the use of lenses? 

By means of the pin4iole disk. In cases where the vision 
is greatly impaired, if the pin-hole produces a decided improve- 
ment in vision, we can assume that the impairment of vision is 
due to ametropia and that an equal or even greater improvement 



152 State Board Examinations 

can be expected from lenses. Whereas if the pin-hole fails to 
improve vision the impairment is probably due to disease and 
glasses would be useless. 

What is the rule for correcting hypermetropia and myopia 
(a) when the patient is fifty years of age ; (b) when the patient is 
eighteen years of age? What is the rule in each case if heterophoria 
also is found? 

(a) When patient is fifty years of age there is little danger 
to fear from the accommodation, and hence the manifest error 
as found could be safely corrected, inclining to the stronger 
convex lenses in hypermetropia and the weaker concave lenses in 
myopia. 

(b) At eighteen years of age in the presence of an active 
accommodation and probably more or less spasm, we cannot 
always give the convex lenses as strong in hypermetropia as 
we might like on account of the objections of the patient, and 
yet even at this age our effort always is to give convex lenses 
as strong and concave lenses as weak as possible. 

Esophoria in connection with hypermetropia would indicate 
a stronger convex lens, while exophoria would call for a weaker 
one. 

Esophoria in connection with myopia would indicate a 
weaker concave lens, and exophoria a stronger one. 



Describe two methods of measuring (a) adduction, (b) abduc- 
tion, (c) sursumduction. State the theoretic value of these tests. 

By prisms from the trial case where they must be removed 
and replaced by stronger ones, and by the rotary prism where 
the strength can be gradually increased without removal from 
before the eye. 

(a) Adduction is measured by the strongest prisms bases 
out which the eyes are able to overcome. 

(b) Abduction by the strongest prisms bases in. 

(c) Sursumduction by the strongest prisms bases up or 
down. 

Theoretically these tests are supposed to show the full 
power of the different muscles. 



Theoretic Optometry 153 

With vision of 20 20 what conditions of refraction may exists 
Emmetropia or slight h}-permetropia or slight hypermetropic 



With vision of 20 20 what kinds of ametropia are eliminated? 
Myopia and high degrees of hypermetropia and astigmatism. 



If vision at 20 feet is improved hy a — .50 D. lens what refrac- 
tive conditions may exist? 

There may be a slight degree of actual myopia or the refrac- 
tion may be emmetropic or hypermetropic, in which a spasm 
of accommodation will cause a concave lens to be accepted. 



A hyperope with exophoria aged twenty has 20/20 vision with 
-\- 1 D. Is this the proper correction? 

We are not told whether this + 1 D. is the strongest convex 
lens that would be accepted, but even if it is, in the presence of 
exophoria, it probably does not represent the full amount of the 
hypermetropia, because the convergence is under strain to over- 
come the tendency to divergence. In order to measure the 
full refractive error there must be relaxation of both convergence 
and accommodation, and this can be accomplished by combining 
prisms bases in with the convex lenses, when it will probably 
be found that a stronger than -f 1 D. will be accepted. 



With a prism base down before the left eye the upper image 
is to the left of the lower image. What name is given to the con- 
dition? Why does the eye deviate? Where should the base of the 
correcting prism be placed? 

The condition is esophoria and diplopia homonymous. The 
eye deviates inward either from excess of convergence or weak- 
ness of divergence. The correcting prism is placed base out. 



What is the amplitude of accommodation of a hypermetrope 
of 2 D. whose near point is at 10 inches? 



154 State Board Examinations 

The emmetropic eye in order to see at 10 inches will use 4 D. 
of accommodation, but here there is an additional 2 D. required 
to overcome the hypermetropia, hence in this particular case 
the amplitude of accommodation is 6 D. 



If a hypermetrope of 2 D, has his near point of 16 inches, 
what is the amount of the presbyopia and what is the correction for 
near use? 

The near point of 16 inches shows an amplitude of accom- 
modation of 2.50 D. For comfortable reading the person should 
possess 5 D. of accommodation; hence there is a deficiency of 
2.50 D. We are told there is a hypermetropia of 2 D., and then 
the other .50 D. must be due to presbyopia of this amount. 

The correction for near use would be + 2.50 D., covering 
both the hypermetropia and the presbyopia. 



We are told to subtract the far point from the near point for 
the amplitude of accommodation in myopia, and to add them 
together in hypermetropia. What is the explanation of this rule? 

In myopia the amplitude of accommodation seems to be 
greater on account of the excess of refractive power of the 
myopic eye; hence in order to find the actual amplitude of 
accommodation we must subtract the amount of myopia. 

In hypermetropia the amplitude of accommodation seems 
to be diminished on account of the deficiency of refractive power 
of the hypermetropic eye ; hence in order to find the real amplitude 
of accommodation we must add the amount that is used to over- 
come the hypermetropia. 

Which tint of glass is best for eyes that are bothered by sunlight? 

There is a difference of opinion on this point. In the majority 
of cases smoked glasses will afford the most relief, and they have 
the advantage of not producing any false color effects. In a 
certain class of cases blue glasses are to be preferred, because 
they absorb the irritating and stimulating red rays. Of late 
years amber glasses have been recommended, because they 



Theoretic Optometry 155 

suppress the chemical rays and, while softening dazzling reflec- 
tions, do not diminish the amount of light admitted to the eye. 
The very latest, and perhaps the best tinted lenses are Crookes. 



Explain why in cases of ametropia the pin-hole disk always 
improves vision. 

If a lighted candle be placed in front of a thin metallic plate 
having a small aperture or pin-hole, an image of the flame will 
be formed upon a screen of white paper or cardboard that is 
placed back of it. Most of the rays proceeding from the candle 
will be intercepted by the plate, but one ray from each point of 
the candle will pass through the pin-hole, and as they pass they 
cross and form an inverted image. The ray from the tip of the 
flame forms the tip of the inverted image, and the ray from 
the bottom of the candle forms the top of the image. A camera 
can be constructed upon this principle and photographs made 
without the use of a lens. 

When impaired vision is due to the imperfect focusing of 
an ametropic eye the pin-hole disk, by cutting off so many of 
the rays, nullifies the refractive power of the eye, and the retinal 
image is formed by the pin-hole upon the principles just described. 

The pin-hole in like manner destroys the effect of any lens, 
no matter how strong, a fact that can be easily demonstrated. 
Take a moderately strong convex lens and hold it close to the 
eye, when the letters on the test card across the room will be 
hardly discernible or entirely blotted out. As soon as the pin- 
hole disk is imposed the refractive power of the lens is destroyed 
and the letters are as clearly seen as the naked vision of the eye 
will allow. 



If you had a patient forty-five years of age wearing for distance 
0. U. — 1 D. sph., with which his near point is 11 3/7 inches, what 
will he his correction for near use? 

This near point represents an amplitude of accommodation 
of 3.50 D. Inasmuch as this is accomplished through his distance 
glasses of — 1 D. his natural accommodation without glasses 



156 State Board Examinations 

would be 4.50 D. This would probably suffice for comfortable 
reading without assistance, but theoretically with the idea that 
the amplitude of accommodation should not fall below 5 D. a 
+ .50 D. lens would be the proper correction for near use. 



Why is it that we cannot tell from the test with the ophthal- 
mometer the full astigmatic error of the eye? 

Because it measures only the corneal astigmatism and is 
unable to reach the astigmatism that may be located in the 
crystalline lens. 

In using the ophthalmometer the mires are brought into exact 
contact in one principal meridian, and are then found to overlap 
in the other principal meridian; what cylinder is indicated, convex 
or concave, and on what axis? 

If we knew the normal curvature of this particular cornea 
for the first principal meridian where the mires were in exact 
contact then an overlapping would show a myopia in the second 
meridian. Or if this second meridian where the overlapping 
occurred was normal then the first meridian is hypermetropic. 

But as a matter of fact we do not use the ophthalmometer 
to ascertain whether the cylinder should be convex or concave, 
but to measure the difference in power between the two principal 
meridians (which is the amount of astigmatism) and the location 
of these two meridians. 

If a convex cylinder is required its axis will correspond to 
the meridian of overlapping; if a concave cylinder, to the meridian 
which showed exact contact. 



When in the fogging test the lines which are most blurred are 
the vertical lines, in which meridian of the eye is the refractive 
error, the vertical or the horizontal? 

Ordinarily the answer would be in the horizontal meridian, 
but this is true only in case of a hypermetropic astigmatism in 
which the fogging lens caused complete relaxation of the accom- 
modation. 



Theoretic Optometry 157 

It is possible that the horizontal meridian may be emme- 
tropic, and if the ciliary muscle failed to relax In response to the 
fogging lens it would be made artificially myopic and cause the 
vertical lines to appear indistinct, when really the hypermetropic 
error would lie in the vertical meridian. 



In a static eye fully corrected with a — 1 sph. combined with 
— 50 cyl. axis 90 , where will be the neutral points of the two 
principal meridians of the eye? 

This correction would indicate a myopia of 1 D. in the 90th 
meridian, and of 1.50 D. in the 180th meridian; and the neutral 
points would be at 40 inches and 26 inches respectively. 



Convergent light entering the eye comes to an exact focus; 
what kind of an eye is it? What will be the focus of far distant 
objects in such an eye? 

The hypermetropic eye is the only form of eye adapted for 
convergent rays. If the accommodation was at rest the focus 
of parallel rays emanating from far distant objects would be 
behind the retina. 

A person fifty years of age looking at an object 20 inches away 
with — 4 D. sph. ^ -\- 5 D. cyl. axis 90^ , accommodates 1 D. 
What is the reading and distance correction? 

In order to look at an object 20 inches away 2 D. of accom- 
modation must be used, but as this person accommodates only 
1 D. we must conclude that the other 1 D. is supplied by the 
uncorrected myopia. Hence the distance correction would be: 
- 5 D. sph. O + 5 D. cyl. axis 90° 
For near vision we would add about + 2.50 D., which would 
make the reading correction: 

- 2.50 D. sph. O + 5 D. cyl. axis 90° 



Name the several methods of determining the nature of a 
refractive error. 



158 State Board Examinations 

Trial case and test types, retinoscopy, ophthalmoscopy, 
refractometer or optometer, Scheiner's test, chromatic test, 
measuring amplitude of accommodation and stenopaic slit. 



In looking through a prism which way does the eye turn and 
in which direction is the object seemingly displaced? 

The eye turns toward the apex of the prism and the object 
is apparently displaced in the same direction. 



In a certain lens the eye first looks through a point 1 mm. 
above the optical center of the lens and then through a point 1 mm. 
to the side of the optical center. The deviation of a distant object 
is greater in the first case than in the second. What kind of a lens 
is it? 

Since there is greater deviation when looking upward than 
when looking to the side there must be greater curvature in the 
vertical meridian than in the horizontal. This would indicate 
a convex sphero-cylindrical lens, with the axis of the convex 
cylinder horizontal. 



Describe the optical principles involved in placing prisms before 
the eyes for the purpose of ascertaining the powers of abduction 
and adduction? 

In orthophoric eyes the image is assumed to fall on each 
macula without muscular effort. When a prism is placed before 
one or both eyes, the rays of light are turned out of their course 
and bent toward the base of the prism. Under such circumstances 
the image would fall on the retina away from the macula and 
produce diplopia, except that the eye as far as it is able in- 
voluntarily turns toward the apex of the prism in order to keep 
the image on the macula and prevent diplopia. 

Therefore, when the prism is placed base in, the eye turns 
out in compensation, and the strongest prism base in which the 
external recti muscles are able to overcome and maintain bin- 
ocular vision will represent the power of abduction. 



Theoretic Optometry 159 

Contrariwise, when the prism is placed base out, the eye 
involuntarily turns in as far as it is able, and the strongest prism 
base out which the internal recti muscles are able to overcome 
and maintain binocular vision will represent the power of adduc- 
tion. 



A hypermetrope of 2 D. has an amplitude of accommodation 
of 4 D. Make the calculation for lenses to he used in reading at 
33 cm. so that one-fourth of the accommodation may he held in 
reserve. 

If it is desired to hold in reserve one-fourth of this 4 D. of 
amplitude of accommodation, there will be only 3 D. of accom- 
modation available for use, of which 2 D. will be taken to over- 
come the hypermetropia, leaving 1 D. for use in reading. But 
as 3 D. is required for use at di?) cm., the accommodation must 
be suplemented by a + 2 D. lens. 



What is the practical value of the pin-hole disk and of the 
stenopaic disk? 

The value of the pin-hole disk is to determine in any case of 
greatly impaired vision whether it is due to an error of refraction 
that can be corrected by lenses or to a diseased condition that 
is beyond optical assistance. 

Of the stenopaic disk to locate the two chief meridians of 
the eye, that of best and of poorest vision, and by means of lenses 
placed before it to measure the refraction of those two meridians. 



A patient looks at a light 20 feet away and a prism strong enough 
to produce douhle vision is placed hase up hefore the right eye; state 
the defect and its amount when the lower light is located to the left and 
a prism of 4° hefore one eye is required to place hoth lights in a vertical 
line, and what will he the position of the 4° prism? 

If prism is base up before right eye, the lower image would 
belong to this eye, and if located to the left would indicate exo- 
phoria, to the extent of the correcting prism which is placed base in. 



160 State Board Examinations 

On which side of the n^iedian line will the streak of light pro- 
duced hy the Maddox rod before the right eye appear in esophoria. 

Right side. 



What is meant by the term refraction? And what is meant by 
the term refraction of the eye? 

Refraction means the bending of rays of Hght as they pass 
obhquely from one medium into another of different density. 

Refraction of the eye is the action of the dioptric media of 
the eye on the Hght that enters it in bringing it to a focus either 
on or off the retina. 

It signifies the optical condition of the eye, whether emme- 
tropic, hypermetropic or myopic. 



The refraction of the vertical meridian is 1.50 D. hypermetropic 
and of the horizontal meridian .25 D. myopic. What is the prescrip- 
tion required? 

+ 1.50 D. cyl. axis 180° O - .25 D. cyl. axis 90°, 
or, 

+ 1.50 D. sph. O - 1.75 D. cyl. axis 90°. 



What is the refractive condition of an eye which requires a 
+ 1.25 cyl. axis 60° for reading and a — 1.25 cyl. axis 150° for 
distance? 

Simple myopic astigmatism of 1.25, in connection with 
presbyopia of like amount. 



The prescription for a certain eye is — .50 Z^ — .25 cyl. axis 
75°; what will be the dioptric power of the eye in its two principal 
meridians, assuming that 58 D. represents emmetropia? \ 

As myopia acts in the way of increasing the refractive power 
of the eye, there would be 58.50 D. in the 75th meridian and 58.75 
D. in the 165th meridian. 






Theoretic Optometry 



161 



Give some description of the Risky Prism and how it is used? 

This consists of two 15° prisms mounted in two cells, placed 
in apposition and controlled by a milled head screw. When so 
placed that the apex of one prism lies against the base of the 
other, their prismatic power is neutralized, and their only effect 
is that of a thick piece of glass. As the milled head is turned and 
the prisms are rotated away from the position of neutralization, 
the effect of a single increasing prism is produced, which reaches 
its greatest power (30°) when the two bases coincide. 

When the line on the prism points to zero on the scale in the 
vertical position, a horizontal prismatic power can be developed 
by rotation. When the line points to zero in the horizontal posi- 
tion, vertical prismatic power can be developed by rotation. 

In any of these cases the amount of power base in, out, up or 
down, can be read off the scale. 

The Risley prism can be used to measure the duction power of 
the various muscles. 



The 180th meridian of one eye is 5 D. hypermetropic, and the 
90th meridian 6 D. hypermetropic. If torics are prescribed, the 
convex curve of the same being -\- 8 D. sphere, what will be the dioptric 
power of the concave surface in the two principal meridians? 

This can, perhaps, be best worked out and understood by 
diagrams. 

The first diagram shows the desired power in each meridian. 

+ 6 D. 





+ 8 D. 
- 2 D. 




+ 6 D. 




+ 8 D. 
-3D. 




+ 5D. 



+ 5 D. 



The second diagram shows in order to maintain these powers 
unaltered, how much concavity must be added in each meridian, 
viz.: — 2 D. in the vertical meridian and — 3 D. in the horizontal 
meridian. 



162 State Board Examinations 

If a certain eye has a power in the 90th meridian of 54 D. and 
in the 180th meridian 53 D., what sphero cylinder must be prescribed 
to give emmetropia for distance, if, with the accommodation at rest, 
a perfect retinal image required the eye to have a dioptric power of 
53.50 D? 

If a dioptric power of 53.5 D. is assumed to represent emme- 
tropia, then the vertical meridian which has a power of 54 D. 
, shows an excess of dioptric power of .50 D., which means a myopia 
of Hke amount; and the horizontal meridian which has a power 
of .53 D. shows a deficiency of dioptric pow^er of .50 D., which 
means a hypermetropia of like amount. 

To correct this condition we must place a — .50 D. in vertical 
meridian to reduce it to 53.50 D., and a + -50 D. in horizontal 
meridian to raise it to 53.50 D., as follows: — .50 D. cyl. axis 180° 
O + .50 D. cyl. axis 90°, which is transposed to — .50 D sphere 
O + 1 D. cyl. axis 90°. 



When the vertical lines on a clock dial are clear, and the hori- 
zontal are not, which meridian will require a concave cylinder to 
correct it? 

The rule is to place the axis of the correcting cylinder in the 
same direction as the indistinct lines, which in this case would be 
horizontal. 

In seeking the explanation of this, it must be remembered | 
that the horizontal lines are seen by the vertical meridian of theJ 
eye and as they are indistinct in this case it must be because the] 
vertical meridian is ametropic; hence we place the axis of the] 
cylinder at 180° in order to bring the refractive power of the lens 
over the vertical meridian, which is the one calling for correction.: 



How is it possible to know with certainty that an eye when being 
tested has been brought out of the fog or not? 

It is safe to assume the eye is out of the fog when the acute- 
ness of vision has been raised to 20/20. However, it may then be 
too much out of the fog, which is something we want to avoid, 
and, therefore, we will try an extra + .25 or -\- .50 and see if it 



Theoretic Optometry 163 

blurs and how much. Or better still, we repeat the whole proce- 
dure, taking care to stop as soon as fairly normal vision is attained. 



Which hlurs the most, objects as seen by a hypermetrope, or as 
seen by a myope, and why? 

This depends on the degree of ametropia and whether refer- 
ence is made to distant or near objects. With equal degrees of 
defect, vision for distant objects would be more blurred in myopia, 
because the hypermetrope is able by means of his accommodation 
to neutralize his deficiency of refractive power and make vision 
normal. 

Whereas in myopia the use of the accommodation (which is 
the only function we possess to influence the distinctness of the 
retinal image) would make the blurred vision still more indistinct. 

But there are some cases of high hypermetropia in which 
vision is very much impaired, and the accommodation is unable 
to raise it to normal, probably on account of the undeveloped 
condition of the eyeball. 

What is meant by the term ''meridian of greatest refraction,'' 
and how is it known? 

In regular astigmatism there are two principal meridians, one 
of greatest and one of least curvature at right angles to each other. 
In myopic astigmatism the meridian of greatest refraction is that 
meridian which show^s the most myopia, and in hypermetropic 
astigmatism the one showing least hypermetropia. In the first 
case it is the meridian right angles to the axis of the concave 
cylinder; and in the second the meridian that corresponds to the 
axis of the convex cylinder. 

It may be detected by the fogging method used in two ways: 

Fog first a set of vertical lines and then a set of horizontal 
lines, and the meridian of greatest refractive power would cor- 
respond to the direction of the lines that can be seen with the 
strongest convex sphere. 

Or fogging the w^hole astigmatic dial, it is in the same direc- 
tion as the lines that first come out of the fog. 

Or it can be found by the retinoscope which measures the 
refraction of each meridian in turn. 



164 State Board Examinations 

When are parallel lines as shown on astigmatic charts reliable, 
and when not? 

These parallel lines radiate in all directions, and they are to 
be relied upon as a test for astigmatism only when the accommo- 
dation is at rest (as in fogging), as otherwise the patient would 
instinctively make an effort to clear the lines, and thus mask the 
astigmatism or lead to an incorrect diagnosis. 



Which should he corrected first in rhinoscopy, the spherical 
error or the astigmatic error, and what is the reason? 

The writer thinks it is best to correct first the meridian of 
least refraction, and then the meridian of greatest refraction. The 
first w^ould represent the spherical error and the difference between 
the two the astigmatic error. 



When should a person with high astigmatism get normal vision 
with his correction, and when would this not he true? 

Sometimes it is impossible in high astigmatism to afford nor- 
mal acuteness of vision, and this is especially true if the error is 
not corrected until later in life. If, however, cylindrical lenses 
had been worn from an early age, and they needed changing as 
they usually do from time to time, we would expect the new 
lenses to give fair vision if there is not too much difference in the 
strength of the lenses or the position of the axis from the previous 
ones. 

Where glasses had not been previously worn and the sensi- 
bility of the retina was blunted by the imperfect images focused 
upon it, vision with the cylinders will be more or less unsatisfac- 
tory, although we would expect it to improve with the constant 
wearing of the glasses. 



How is the test for adduction usually made? 

By prisms bases out gradually increased in strength. The 
strongest prisms with which single vision of a light can be main- 
tained represent the power of adduction. This is a variable quan- 



Theoretic Optometry 165 

tity and ranges from 15° to 40° or higher, and can be greatly 
increased bv exercise. 



When two phis lenses of different power give equally good 
vision, why is the stronger one selected? 

We presume that this has reference to the tests for distant 
vision in hypermetropia, in which the rule is to select the strong- 
est convex lens the patient can be made to accept, for the reason 
that the stronger the lens, the more of the hypermetropia it cor- 
rects and the less effort on the part of the accommodation is 
required. 

If a patient has astigmatism which requires a -\- 2 D. cylinder 
axis horizontal to correct, and you place a -\- 2 D. sphere before the 
eye, what artificial condition is produced? 

The + 2 D. sphere corrects the hypermetropic meridian, but 
at the same time it blurs the emmetropic meridian and makes it 
myopic to the extent of 2 D. Therefore, the artificial condition 
produced is simply myopic astigmatism, at right angles to the 
original hypermetropic astigmatism. 



What is the principal purpose of each of the following instru- 
ments: ophthalmoscope, retinoscope and ophthalmometer? 

The ophthalmoscope, to examine the interior of the eyeball 
as to the presence or absence of disease, and if disease is present 
to determine its location and nature. 

The retinoscope, to measure the refractive condition of the 
eye. 

The ophthalmometer, to detect and measure corneal astig- 
matism as to its amount and the location of the two principal 
meridians. 

Under what conditions may a person have an amplitude of 
accommodation of 3 D. and still have no range of vision? 

When hypermetropia is present to an equal or greater 
amount. 



166 State Board Examinations 

If a patient has myopic astigmatism, under what conditions 
would a plus cylinder he proper to prescribe? 

For reading after presbyopia has set in. For instance, if a 
patient wearing —ID. cyl. axis 180° reaches 45 years of age, 
when the addition of + 1 D. is needed for reading, the prescrip- 
tion would be + 1 D. cyl. axis 90°. 



// light coming from a distance of 20 inches is exactly focused 
on the retina with a -\- 3 D. lens and with the accommodation at 
rest, what is the kind and amount of refractive error? How do you 
calculate the result? 

Light coming from a distance of 20 inches has such a degree 
of divergence that requires + 2 D. lens to focus it upon the retina 
of an eye with the accommodation at rest. 

If, however, as in this case, a + 3 D. lens is required, then 
we must assume that the extra + 1 D. represents a deficiency in 
the refractive power of the eye of that amount or in other words 
a hypermetropia of 1 D. 

How would you test the strength of any particular ocular muscle 
with prisms? 

By placing apex of prism over such muscle and finding strong- 
est prism with which binocular vision can be maintained. 



What is the main purpose of the ophthalmometer? 

To determine the curvatures of the cornea, the location of 
the two principal meridians and the refractive power of each 
meridian. 



// a person 50 years of age came to you wearing a pair of bifocals 
of which the power was 3.50 D., what error in the distance correction 
would you suspect? 

We assume that + 3.50 D. is meant to indicate the power of 
the segments, in which case we would say the distance was under- 



Theoretic Optometry 167 

corrected; for the reason that at this age the segment scarcely 
exceeds 2 D. or in other words the presbyopia is seldom greater 
than this. Hence the 3.50 D. segments are correcting some of 
the hypermetropia, which should have been provided for in the 
distance lenses. 



A child showing with the objective tests 5 D. of hypermetropia 
sees clearly with + 1.50 D. lenses, while anything stronger blurs 
the letters. What would you give and how would you handle the 
case? 

This would depend somewhat on the age of the child and 
the symptoms of which he complains. As a rule children will 
bear fogging better than adults, therefore, if the child were 
quite young and the symptoms urgent, we would give + 2.50 D. 
for constant wear. 

If the child was older and likely to object to the fogging, 
we would attempt to correct only the manifest hypermetropia 
at first with + 1.50 D. lenses, in the thought that after a while 
some of the latent defect would become manifest when the 
strength of the glasses could be increased. After these second 
glasses had been worn for a time, perhaps another increase could 
be borne, if the symptoms were such as to call for more relief. 



What method would you employ to find the amount of latent 
hypermetropia? Can the latent be estimated from the amount of 
manifest? 

I would employ the fogging method to estimate as far as 
possible the total hypermetropia, and the difference between 
that and the manifest hypermetropia would represent the latent. 

Or if it was not possible to discover any latent error in this 
way, I would correct the manifest w^th the thought that under 
such correction some of the latent would gradually become 
manifest. 

The amount of latent bears no constant relation to the 
amount of manifest error; in youth it is nearly all latent and in 
old age it is nearly all manifest. 



168 State Board Examinations 

If the internal recti muscles were weak, how would the prism 
exercise he given to make them stronger? 

As the apex of the prism is always to be placed over the 
muscle to be exercised, the prisms in this case are to be placed 
bases out. 

The first point is to ascertain the power of the internal recti 
and thus determine how far they fall below the normal standard, 
which is accomplished by means of prisms bases out, the strongest 
with which single vision of a light can be maintained. This may 
fall to 10° or less as compared with the normal power of conver- 
gence, which is from 20° to 30°. 

Having found this strongest prism, it is divided between the 
two eyes and alternately raised and lowered. Each time the 
prisms are lowered the light is probably seen double momentarily, 
but is quickly brought into one by the action of the internal 
recti. The raising of the prisms allows these muscles to relax 
a little and get a slight rest, and their lowering calls the muscles 
into action again. 

This alternate relaxation and contraction stimulates the 
innervation and circulation of the muscles, and it is soon found 
that stronger prisms can be overcome, and hence they are 
gradually strengthened as the increasing power of convergence 
will allow. 



A myope is found to he wearing his spectacles each — 10 D. 
sphere tilted so that the right glass is 5 mm. ahove the center of his 
pupil and the left glass is 5 mm. helow. What will he the effect on 
his vision and how many degrees would it displace the image at 
3 meters? 

According to the rule of decentration that there would be 1° 
prismatic effect for each diopter of refractive power when 
decentered 10 mm., there would be 5° prism power added to the 
concave lens in each eye. In the right eye the prismatic effect 
would be base down, which causes a displacement upward of 
object looked at, and in left eye base up causing displacement 
downward. 

As the vertical muscles are comparatively weak and easily 
thrown out of balance, such tilting of the lenses would almost 



Theoretic Optometry 169 

certainly cause diplopia, unless there was a condition of right 
hyperphoria, which would be corrected by such position of the 
lenses. 

Inasmuch as the above figures apply for each meter of 
distance, the displacement at 3 meters would be three times as 
great. 

A person requires for the right eye alone — 6 D. sphere for 
distance and — 3D. sphere for reading, hut wants only one pair 
of spectacles. State the various ways in which this can he managed, 
and which method you prefer. 

As the question states the glasses are required for the right 
eye alone, we must assume that the left eye is useless as an organ 
of vision, and the effect of the proper glass before the right 
eye can be obtained in one of several ways. 

1. By using a reversible eyeglass frame with regular guards, 
or reversible spectacles with a C or X bridge, and placing a — 6 
D. lens in one of the eyepieces of the frame and a — 3 D. lens 
in the other eye wire, which will allow the lens desired to be 
placed before the right eye at will. 

2. By a frame fitted with — 6 D. lenses and a hook front 
fitted with + 3 D. lenses. 

3. By a frame fitted with — 3D. lenses and a hook front 
fitted with — 3D. lenses. 

4. By giving bifocals, with — 6 D. for distance and + 3 
added for reading. 

What is the effect of having strong lenses placed too far from 
the eye? 

As a strong convex lens is moved away from the eye the 
object seems to approach and is magnified; as a concave lens is 
moved away the object seems to recede and is minified. 



In testing an eye with- the stenopaic slit and using a set of 
parallel test lines as the ohject of regard, how should these lines 
be arranged in reference to the direction of the slit? 

Vertical lines are seen by the horizontal meridian of the 



170 State Board Examinations 

eye and horizontal lines by the vertical meridian ; hence the lines 
should always be at right angles to the direction of the slit. 

But as a matter of fact the card of test letters should be 
used in connection with the stenopaic slit and not the card of 
radiating lines. 



On what principle are the sizes of the letters on the test chart 
determined? 

Inasmuch as a visual angle of one minute is the smallest 
that can be perceived, the test letters are made of such size 
that the width of each line of the letter at its proper distance 
from the eye will form an angle of one minute, w^hile the w^hole 
letter will form an angle of five minutes. 



// a cylinder is placed before the emmetropic eye and two sets 
of lines are looked at, one agreeing with the axis of the cylinder and 
the other at right angles to this, will one set of lines he seen best? 

Assuming the eye to be in a static condition, the lines agreeing 
with the axis of the cylinder will be more or less blurred according 
to the nature and strength of the cylinder, while the lines at 
right angles to this will be unimpaired because seen through 
the axis of the cylinder, which is piano, always bearing in mind 
the fact that vertical lines are seen by horizontal meridian of 
eve, and horizontal lines bv vertical meridian of eve. 



A pinhole disk is usually claimed as showing whether a 
refractive error is present or not. Under what condition would 
this not be correct? 

If in a case of impaired vision the pinhole disk raised it to 
normal, w^e are justified in saying a refractive error is present. 
If, how^ever, there was opacity of any refracting medium or 
atrophy of the optic nerve, there could be no improvement in 
vision on this account, and yet there might be an error of refrac- 
tion in addition. 



Theoretic Optometry 171 

// aw eye is 2 D. hypermetropic in the vertical meridian and 
you place a -\- 2 sphere in front of this eye, what is the condition 
thus artificially produced? 

This + 2 D. sphere will correct the vertical hypermetropic 
meridian and make it emmetropic, but at the same time it will 
make the emmetropic horizontal meridian artificially myopic 
to the extent of 2 D. 



Which set of lines in a cross-dial chart shows the axis of the 
astigmatism? 

The horizontal lines are seen by the vertical meridian of 
the eye, and the vertical lines by the horizontal meridian of the 
eye. Therefore, if the horizontal lines are indistinct, the vertical 
meridian of the eye is assumed to be defective; and if the vertical 
lines are indistinct, we assume that the horizontal meridian of 
the eye is defective. 

It must be further remembered that the meridian of the 
axis of a cylinder is plane and that the refractive power lies in 
the meridian at right angles thereto. Hence we place the axis 
at right angles to the defective meridian; and the indistinct lines 
are also at right angles to the defective meridian, and therefore 
the axis must be in the same meridian as the indistinct lines. 

But suppose the defective meridian was hypermetropic and 
the accommodation came into action, as is likely to be the case; 
then the naturally ametropic meridian is made artificially emme- 
tropic, and the naturally emmetropic meridian is made artificially 
ametropic, and under these circumstances the rule mentioned 
above will not apply at all as the axes are reversed. 

For instance, in a case of simple hypermetropic astigmatism 
with the rule, the proper correcting lens is a + cyl. axis 90°; 
but if the accommodation is allowed to come into play, the lens 
accepted would be a — cyl. axis 180°. 

Therefore, the use of the cross lines to determine the location 
for the axis of the cylinder is of value only when the accommoda- 
tion is kept passive, as in the fogging method when it is repressed 
by a convex lens. 



What is meant by exercise of the muscles or ocular gymnastics? 



172 State Board Examinations 

In cases of heterophoria, after the proper lenses have been 
given to correct the error of refraction, and perhaps the patient 
referred to his family physician to have his general health looked 
after, an effort may be made to strengthen the weak muscles by 
systematic exercise, to which the term "ocular gymnastics" has 
been applied. 

The most promising cases are those of exophoria due to an 
insufficiency of the internal recti, which set of muscles we try 
to strengthen by exercise. 

In the first exercise the patient is directed to take a pencil 
and hold it at arm's length in a perpendicular position. It 
should be held in the median line in front and gradually 
approached while the top of the pencil is fixed by the two eyes. 
When it gets to within a few inches from the e\'es, the pencil 
is seen double, when it should at once be pushed off the starting 
point and the procedure done over again. This exercise can be 
kept up for five or ten minutes (if it does not nauseate the patient, 
as sometimes happens) and repeated daily. 

The second exercise is by means of prisms placed over the 
eyes bases out, commencing with weak numbers and gradually 
increasing their strength while the patient looks at a point of 
light across the room. The strongest prisms are found which 
can be overcome and with which singleness of vision of the light 
can be maintained for a moment or two. and they will represent 
the present power of adduction. 

Then the prisms are raised for a few seconds, and as they 
are lowered again diplopia is evident for an instant, but soon 
disappears. The prisms should be raised and lowered every 
thirty seconds for half a dozen times, when the prisms may be 
increased 2°. If his increase can be easily overcome, the prisms 
should be raised and lowered another half dozen times. 

The idea to be kept in mind is to gradually increase the 
prisms about 2° at a time. The optometrist must use his judg- 
ment and not try to increase them too fast. This point can 
usually be determined by the readiness with which the temporary 
diplopia caused by dropping the prisms into place can be over- 
come. If the two lights run together as quick as a flash, then 
probably stronger prisms can be overcome and should be tried. 



Theoretic Optometry . 173 

At the start, perhaps the eyes cannot overcome more than 
6° or 8°. but as the muscles strengthen under the exercise and 
their innervation is stimulated, the prisms can be increased up 
to 40° or 50°. The exercises should be continued until this 
point is reached, because after the}^ are stopped some of the 
adduction power thus developed must be expected to disappear. 



Practical Optometry 

What is the relation hetiveen prisms marked in degrees and 
prisms marked in prism diopters, and ivhat is the relation between 
the value as marked on prisms and their deviating power? 

In the case of prisms marked in degrees we say they are 
numbered by the refracting angle of the two surfaces of the glass 
where they meet at the edge or apex, founded on the fact that 
there are 360° in a circle and that 1 360th part of a circle would 
represent a prism of 1°. 

In the case of prisms numbered in prism diopters the unit of 
this method of numbering is a prism diopter (which may be 
abbreviated P. D. or A) which will deviate a ray of light just 
1 cm. at a distance of 1 meter, or 2 cm. for 2 meters. 3 cm. for 
3 meters and so on in the same proportion for all distances, no 
matter what the index of refraction of the glass used ma^^ be. 

A comparison of prisms marked in degrees with those num- 
bered by prism diopters will show them to have nearly the same 
value; in fact, the difference is so slight that they can be used 
interchangeably in practice. 

The relation between the number marked on the prism and 
its deviating power is as 2 to 1. That is, the angle of deviation is 
equal to about one-half the angle of the prism, which applies to 
all prisms of ten degrees or less. In prisms of a higher power the 
angle of deviation increases. 



When prisms are ordered in a prescription how would you know 
that the prisms have been ground in the lens? 

The first point to be looked for is a difference in thickness in 
the two opposite sides of the lens, comparing the inner and outer 
edges together, and the upper and lower. 

Then we look to see if the lens shows the deviating effect of 
a prism by holding it in front of the eye at some little distance and 
viewing a straight edge or line across the room. If the prescrip- 

174 



Practical Optometry 175 

tion calls for a lens of refractive power combined with the prism, 
we must be careful to look directly through the optical center of 
the lens, as otherwise the deviating power developed by looking 
through any refractive lens at a point away from its center will 
interfere with our efforts to detect the real prism in the combi- 
nation. Perhaps the better way would be to first neutralize the 
refracting power of the lens and even here we must have a care 
to see that its optical center shall exactly coincide with the optical 
center of the neutralizing lens and then the presence of a prism 
becomes evident by a break in the line, no matter what part of the 
lens we look through. 

Does decentering a sphere make a sphero-prism, or to make a 
sphero-prism must the sphere he on one side and the prism on the 
other? 

The decentration of a sphere makes manifest its prismatic 
power, which then has the value of a sphero-prism, although it is 
only a spherical lens. 

It would be impossible to have a prism on one surface of a 
lens, because a prism depends upon its two plane surfaces being 
inclined to each other, hence one surface of a lens could not show 
the effect of a prism. When a sphero-prism is ordered, a plane 
prism is taken and the desired spherical curvature is ground on 
one of its surfaces, the other surface remaining plane. 



Upon what optical principle is the use of the Maddox rod based? 

The Maddox rod produces very great magnification at right 
angles to its axis, elongating a small flame into a long streak of 
light. This result is such a great dissimilarity in the size, shape and 
appearance of the two retinal images that they cannot be fused 
into one, and as the eyes abandon the attempt to effect fusion, 
each eye is left to the action of its muscles and the eyes assume 
the position of unstrained equilibrium. Under these circumstances 
the position which the streak assumes with reference to the light 
will indicate either orthophoria or heterophoria. 



Which position should he given to the prism placed on the 
Maddox rod hefore the normal right eye so as to deflect the observed 



176 State Board Examinations 

vertical line of light 12 cm. to the left at 6 meters distance? What is 
the power of the prism? 

The displacement of objects viewed through a prism is always 
in the direction of the apex; therefore, in this case where it is 
desired to move the right image inwards or to the left, the prism 
would have to be placed base out over right eye. 

Inasmuch as a prism diopter causes a deviation in a ray of 
light in the ratio of 1 cm. for each meter of distance, there will 
be a deviation of 3 cm. for 3 meters and 6 cm. for 6 meters. In 
this case where the deviation is 12 cm. for 6 meters, the power of 
the prism would be 2 A. 



In which position must the prism bases be placed to ascertain 
the powers of adduction, abduction, supraduction and infraduction, 
respectively? 

Adduction, bases out. 
Abduction, bases in. 
Supraduction, bases down. 
Infraduction, bases up. 



Would you always prescribe glasses where ametropia is found 
to exist? If not, state exceptions. 

No; in slight degrees of defect, where vision was but little 
if any impaired, and where there were no symptoms complained 
of, glasses would scarcely be justified; especially if slight hyper- 
metropia existed with exophoria, or low myopia with esophoria. 



{ 



With the Maddox rod horizontal before the left eye, the point 
of light being at six meters and the streak appearing 12 cm. to the 
right of the point, what is the kind and amount of heterophoria and 
in which direction would you place the base of a prism to bring the 
streak through the point of light? 

As the diplopia in this case is crossed the condition is one 
of exophoria. 



Practical Optormlry 177 

The de\-iation of one prism diopter is 1 cm. for each meter 
of distance, which would mean 6 cm. for 6m. Therefore, a dis- 
placement of 12 cm. would indicate 2 A of exophoria. 

The base of prism is placed in to bring the streak through 
the light. 



A plus 4 diopter lens is the correction called for in a case of 
hypermetropia. What is the effect of wearing this lens too far from 
the eye, say an inch, as, for instance, where the spectacles are pushed 
far down on the nose? 

The effect of pushing the lens forward is to increase its 
strength. Therefore if + 4 D. is the proper correction in the usual 
position, it would be too strong when worn far down the nose. 



If yays of light from a distance of 20 inches are focused on 
the retina with a plus 2 D. lens without accommodation, what is the 
refractive error? 

Rays of light diverging from a distance of 20 inches and 
passing through a convex lens of 2 D. would emerge parallel, 
and if such rays were focused on the retina without accommo- 
dation, the eye must be emmetropic. 



What is the ametropic correction for a patient, age twenty 
years, with his near point of the vertical meridian at 8 inches and 
the near point of the horizontal meridian at 10 inches? 

The emmetropic eye at 20 years of age possesses 10 D. 
amplitude of accommodation in both meridians. 

In this case a near point of 8 inches in the vertical meridian 
shows 5 D. of accommodation for this meridian, which is a 
deficiency of 5 D. and a probable hypermetropia of this amount. 
In the horizontal meridian the near point of 10 inches shows 4 D. 
of accommodation on a deficiency of 6 D. and a probable hyper- 
metropia of this amount. This would be compound hyper- 
metropic astigmatism and the correction reads as follows: 
+ 5 D. sph. O + 1 D. cyl. axis 90° 



178 "State Board Examinations 

What is the correct prescription for distance, for a patient 
aged thirty years, looking at 13 inches with his focus on the retina, 
while wearing a -{- 2 sphere on a — 3 cyl. axis 90°, using 2 D. of 
accommodation? 

An emmetropic eye looking at 13 inches uses 3 D. of accom- 
modation. If, however, in this case a -|- 2 D. sphere is worn and 
in addition it is necessary to use 2 D. of accommodation, then 
this 4 D. would indicate that 1 D. more than in emmetropia 
is necessary, revealing a hypermetropia of this amount; therefore 
the correct prescription for distance would be 

+ 1 D. sph. C - 3 D. cyl. axis 90° 



A patient is told to look at a small light 20 feet away and is 
given diplopia with a prism, base down, before the right eye. What 
defect and amount of error are present when the upper light is 
located to the right, and a prism of 3° before each eye is required 
to bring both lights to the median line? What is the position of the 
3° prisms? 

If the prism is base down before right eye then the right 
image belongs to right eye and the condition is one of esophoria. 
The correcting prisms of 3° must be placed bases out to bring the 
lights one under the other. 



A myope of 3 D. sees with his right eye the Maddox rod line 
9 cm. lower than the object light. Write the prescription for distance 
lenses, lenses to be eye, so that the least cost will be incurred. 

The deviation caused by one prism diopter is 1 cm. for each 
meter distance. Presuming this test was made at the usual dis- 
tance of 6 meters, the deviation would equal 6 cm. The question 
says there is a deviation of 9 cm., which, figured out on the above 
basis, would correspond to 1>2 prism D. 

The amount of prismatic power that can be developed by 
decentration is on the basis of 1 A for every 1 D. when de- 
centered 10 mm., or >^ A for every 1 D. when decentered 5 mm., 
or 1 W A for this 3 D. lens when decentered 5 mm. 



\ 



Practical Optometry 179 

The rough lens would not be large enough to permit of this 
decentration in the horizontal meridian, but it can be done in 
the vertical meridian, which is 9 mm. less than the horizontal. 

As the right eye image is lower than the left the case is one 
of right hyperphoria, which is to be corrected by prism base 
down right eye, or base up left eye. 

As this is a concave lens the base of prism is opposite to 
the direction of decentration, as the preferable direction for base 
of prism is up, as it is custorriary to place the prism over the left 
eye (unless the right eye is decidedly the poorer eye), and as it 
is desired to avoid prisms in order to incur the least cost, we will 
order 

O. D. - 3 D. sph. 

O. S. — 3 D. sph. decentered down 5 mm. 



What are the points of resemblance and difference between a 
case of spasm of the accommodation and a case of myopia? In 
what kind and what degree of ametropia does spasm of the accom- 
modation generally occur? 

The points of resemblance are that in both conditions the 
reading point is closer than normal, and the acuteness of vision 
is impaired, which is improved by concave lenses. 

The points of difference are: 

(1) In spasm the distance of the far point does not corre- 
spond to the degree of defect as it does in myopia. 

(2) In spasm the visual acuity is constantly varying, while 
in myopia it is more likely to be a fixed quantity either with the 
naked eye or with the correcting lenses. 

(3) In spasm there are complaints of pain and inability 
to use the eyes continuously for near work, while in myopia 
there are no asthenopic symptoms. 

(4) In spasm the near point is apt to be closer to the eyes 
for a given degree of false myopia than for a similar degree of 
real myopia. 

(5) In spasm the distant vision is better than the near, 
while in myopia the near vision is better than the distant. 



What is presbyopia due to and how does it differ from acquired 
hypermetropia? 



180 State Board Examinations 

Presbyopia is due to an inability to accommodate sufficiently 
to make near vision clear, on account of the sclerosis of the 
crystalline lens which comes on with age and becomes noticeable 
in the early 40's only when near vision is attempted. 

As the presbyopic changes progress there comes a time late 
in life when distant vision also becomes impaired as a result of 
the lessened refractive power of the crystalline and then convex 
lenses are needed not only for reading but also for distance, the 
latter being weaker than the former. This condition is known 
as acquired hypermetropia. 



What is meant hy mixed astigmatism? Describe the best 
method of testing such a case and what directions you would give as 
to the wearing of glasses. 

Mixed astigmatism is that condition in which the focal line 
of one meridian lies in front of the retina and of the other meridian 
back of the retina; or in other words, one meridian is myopic 
and the other hypermetropic. 

The best method of subjective testing is (after having used 
the ophthalmometer and determining the presence of astigma- 
tism and the location of the principal meridians) to correct the 
hypermetropic meridian by means of convex cylinders, which 
should be crowded on to the limit of acceptance and w^hich will 
not afTect the other meridian, and then use concave cylinders over 
the convex cylinder with their axes at right angles. The convex 
cylinder will advance the focus of the hypermetropic meridian 
to the retina, and the concave cylinder will throw the focus of 
the myopic meridian back to the retina, so that the image will 
now be composed of focal points on the retina instead of focal 
lines in front and back of it. 

The cross cylinder thus obtained can usually with advantage 
be transposed into a sphero-cylinder, in which case it is sometimes 
desirable to slightly reduce the convex element of the combination 
in order to make them more acceptable to the patient. 

The same principle can be used in the retinoscopic test. 
After having determined the directions of the two principal merid- 
ians the hypermetropic is neutralized by a convex sphere and the 
myopic meridian by a concave sphere, from which, after making 
the necessary allowances, the correcting combination can be de- 



Practical Optometry 181 

duced. remembering that the axis must be placed at right angles 
to the meridian. 

In mixed astigmatism the glasses should be worn constantly 
and care should be taken that the glasses are kept in proper 
adjustment and without displacement of the axes, which in the 
higher degrees of mixed astigmatism would result in impairment 
of vision and asthenopic symptoms. 



If a patient has myopic astigmatism, in what case would plus 
cylinder he prescribed? 

In a case of simple myopic astigmatism, convex cylinders 
could not be prescribed unless combined with the same number 
concave sphere, but this would increase the cost. 

In compound myopic astigmatism the transposition may be 
made in order to get a higher concave surface and thus make a 
periscopic effect, or to change the axis from horizontal to vertical, 
or for both purposes. 

For instance: 

-3D. sph. C - 2 D. cyl. axis 180° 
may be transposed to 

- 5 D. sph. C + 2 D. cyl. axis 90°, 
with the gain of both of the above-mentioned advantages. 



// a person requires plus 5 D. in the vertical meridian and plus 

4 D. in the horizontal and is tested with a small astigmatic clock 
dial, what would he the farthest distance at which the vertical line 
could he distinctly seen; also the horizontal lines, with a plus 8 D. 
lens? 

If the vertical meridian is hypermetropic to the extent of 

5 D. and a + 8 D. lens is placed before it, there is an overcor- 
rection of 3 D., thus producing an artificial myopia of like amount, 
with a far point for this meridian of 13 inches. And as the hori- 
zontal lines are seen by the vertical meridian of the eye, therefore, 
13 inches would be the farthest distance at which the horizontal 
lines could be distinctly seen. 

If the horizontal meridian is hypermetropic 4 D., and a + 8 
D. lens is placed before it, there is an overcorrection of 4 D. with 



182 State Board Examinations 

an artificial myopia of like amount and a far point of 10 inches. 
And as the vertical lines are seen by the horizontal meridian 10 
inches would be the farthest point at which the vertical lines 
could be seen. 

What change in adjustment must he made when a myope looks 
through a pair of field glasses just used hy an emmetrope? 

The tube must be shortened in order to make the emergent 
rays divergent and thus be able to focus on the retina of the 
myopic eye. 

What is the prismatic effect in a pair of glasses set in a frame 
5 mm. too wide when the prescription is as follows: 
R. E. + 2.50 Z: - 1 cyl. axis 90 
L. E. + 3.50 C - 1.50 cyl. axis 90 

The prismatic effect is produced by a decentration in the 
horizontal meridian, the power of which is reduced in both lenses 
by the concave cylinders to 1.50 D. in right and 2 D. in left eye. 

The rule is lA of prismatic power for every diopter when 
decentered 10 mm. In this case where the frame is 5 mm. too 
wide there is 2.5 mm. decentration for each eye, which according 
to the above rule would yield ^A for right eye and ^A for left 
eye, or a total for the two eyes of ^ A of prismatic effect. 



The following is a prescription in which the correction for the 
right eye is omitted. Supply it: 

Distance R. E. - 0.50 - 1.00 cyl. axis 45°. 
Distance L. E. missing. 
Near R. E. + 0.50 C +1.00 cyl. axis 135. 
Near L. E. + 3.50 C — 0.75 cyl. axis 135. 

A comparison of the distance and near corrections of the 
right eye shows the addition of a + 2 D. and a transposition of 
the cylinder from concave to convex. 

Assuming that the same addition was made in the near glass 
of the left eye, we must deduct this 2D., which would leave 

+ 1.50 D. S. - .75 D. cyl. axis 135° 
or by transposition in order to correspond with right eye 
-f .75 D. S. + .75 D. cyl. axis 45° 



Practical Optometry 183 

What advantage is secured by using the Maddox rod test as a 
phorometric test? 

The principal advantage of the Maddox rod Is that Its Image, 
although distorted and elongated into a streak, Is formed upon 
the macula; whereas In the prism test the Images are displaced 
from the macula. 



What errors of refraction may he congenital and which ones 
acquired? 

Usually we regard hypermetropic errors as congenital and 
myopic as acquired. 

What is asthenopia? 

The word means weak sight and Is used in connection with a 
condition where the eyes can be used but for a short time, and Is 
accompanied by more or less discomfort, due to strain of either 
the accommodation or convergence. 



Describe a test of the extrinsic muscles which is dynamic; that 
is, the test must be made with convergence in full action. 

Ask the person being examined to look at the point of a pencil 
which Is held In the median line a short distance in front of the 
face and gradually moved nearer to the eyes until one of them 
is seen to abandon the effort of convergence and deviate from 
the fixation point and turn outward. The position of the pencil 
when this occurs indicates the near point of convergence. This 
has been estimated to be normally about four inches from 'eyes. 



Describe a test of the extrinsic muscles of the eyes that is not 
dynamic; that is, the test is not made with convergence in full force. 

Any test that is made with a distant point of fixation, as for 
instance the Maddox rod test with the light 20 feet away. This 
is a test that Is "not dynamic" because there Is no impulse to the 
convergence for the maintenance of binocular vision. 



184 State Board Examinations 

In an astigmatic eye where would be the circle of least diffusion? 

In astigmatism the two principal meridians each have a 
different focus, the space between which is known as the focal 
interval of Sturm. Xo matter what the position of the retina 
in such an eye, no distinct image can be formed upon it. 

If it is in focus for one meridian it is very much out of focus 
for the other. The accommodation is brought into action to lessen 
as much as possible the diffusion circles, but there can be no true 
focus for both meridians at the same time, and therefore the image 
cannot be sharp and distinct. Probably the best vision or least 
diffusion would be midway between the foci of the two meridians. 



Why are prisms placed bases in and out, respectively, in 
convergent and divergent strabismus? 

Ordinarily, strabismus is beyond the reach of prisms, but 
if correctible by prisms the rule is to place the base opposite to 
deviation, that is, in convergent strabismus, base out, and in 
divergent strabismus, base in. These are the positions of relief 
and assistance. 

But if it is desired to develop and strengthen a set of muscles 
by prism exercise, then the apex of the prism is placed over the 
muscles which it is desired to act upon, that is, in the positions 
named in the question. In convergent strabismus, where it is 
desired to exercise the external recti, the prism is placed apex 
out or base in; in divergent strabismus, where the internal recti 
need toning up, the prism is placed apex in or base out. 



In a young person with 1 diopter of hyperopia the pinhole 
disk makes the letters on the chart worse than when seen with the 
naked eye, while in a myope of 1 diopter there will be an improve- 
ment with this test. What is the reason? 

The pinhole test is one that is of value only when vision is 
impaired. In the case of a young person with 1 D. of hyper- 
metropia the visual acuity is not lessened because the accom- 
modation instinctively neutralizes the deficient refraction. As 
the vision is already normal the pinhole disk cannot make it 



Practical Optometry 185 

any better, but rather makes it worse because it cuts off so 
much Hght that would otherwise enter the eye. 

Whereas, in the case of a myope of 1 D. the acuteness of 
vision is much impaired (probably one-half normal), but it is at 
once raised to normal by the pinhole disk, for the following reason ; 

The pinhole allows the passage of a very narrow pencil of 
light and to that extent diminishes the size of the diffusion circles 
on the retina. If the pinhole be 1/5 the size of the pupil, the 
dift'usion circle on the retina will be only 1/5 the size of the 
usual diffusion circle. 

Suppose the diffusion circle upon the retina covered an area 
of 100 cones; with a pinhole 1/5 the diameter of the pupil the 
area will be reduced 1/25, and only 4 cones be covered; this will 
result in an enormous increase in the visual acuity. 



In a case of hypermetropia of 6 diopters with little accommoda- 
tion how would we measure the amount of facultative hypermetropia? 

By the proportion of the total hypermetropia that can be 
overcome by the accommodation. If this case of 6 D. of hyper- 
metropia was fifty years of age he would possess about 2.50 D. 
of accommodation, which would be used as far as it would go 
in the correction of the defect, and the balance must be corrected 
by lenses. 

In this case there would be 2.50 D. of facultative and 3.50 
D. of absolute hypermetropia. The facultative is the difference 
betw^een the total defect and the w^eakest lens that raises vision 
to normal. 



How would you get the amplitude of the convergence-accom- 
modative function? 

By measuring the closest point at which it is possible to see 
clearly and at the same time that binocular vision can be main- 
tained. The accommodation is expressed in diopters and the 
convergence in meter angles. 

In the emmetropic eye an object situated at 8 inches would 
call for 5 D. of accommodation and 5 meter angles of convergence. 



186 State Board Examinations 

In what way can eye with irregular astigmatism be best cor- 
rected? 

The center of the cornea may be flatter or more convex, 
thus giving it a different refractive power from the periphery; 
or the cornea may be studded with numerous facets, the results of 
previous ulcers; in either case vision is impaired and objects 
distorted. 

It is obvious that such a condition is not correctible by the 
usual forms of lenses. The only method of improving vision 
is to cut off the peripheral rays and allow light to pass through 
a small portion of the cornea, where the curvature is most even 
and the transparency greatest. This can be accomplished by an 
opaque diaphragm with a small opening placed before the cornea 
in the position that is found the most favorable. 



The distance correction is 

O. D. -\- 2 D. sph. O. S. - 1 D. sph. 

What decentration must be given to the added wafers of + 2.50 D. 
for reading in order to neutralize the prismatic effects which would 
otherwise be produced by looking through the centers of the lower 
fields, 10 mm. below the centers of the distance lenses? 

On account of the size of the finished lenses it would be 
utterly impossible to obtain a decentration of 10 mm.; with this 
proviso we will answer the question as it is given. 

The decentration for the two wafers will not be the same 
because the distance lenses vary in character and in power. 

According to the rule which has already been given on these 
pages, the prism strength that would be developed by a down 
decentration of 10 mm. of this convex lens of 2 D. would be 2° 
with base up. Therefore, in order to overcome this the wafer on 
this lens w^ould have to be decentered down to a corresponding 
degree, but as the power of this wafer is 2.50 D. the decentration 
necessary to produce a 2° prismatic value base down to balance 
would be only 8 mm. 

In regard to the left lens, which is a concave lens of 1 D. 
power the decentration downward of 10 mm. w^ould produce a 
prism value of 1°, but in this case with base down. In order to 



Practical Optometry 187 

counterbalance this the decentration of the + 2.50 D. wafer 
would have to be up to the extent of 4 mm. 

In a case of this kind it would be well also to order the 
segments decentered inwards, or rather order the pupillary 
distance of the segments to less than that of the distance lenses, 
to allow for the necessary convergence and decrease of pupillary 
distance when the eyes are used at close range. 



The wearer of a pair of strong mimes lenses complains that 
distant objects appear clear but diminished in size; how may this 
defect be rectified in part at least without altering the power of the 
lenses? 

By pushing the glasses as close to the eyes as possible. 



Describe an accurate subjective method for detecting astigmia 
and determining its character. 

Astigmatism may be detected subjectively in two ways: 

1. Use of the card of radiating lines, and the reply of the 
patient that lines running in one direction are notably plainer 
and more distinct than lines running in some other directions. 

2. Use of the card of test letters and the acceptance by the 
patient of the cylinder as affording better vision than a sphere. 

The character of the astigmatism is determined in each case 
by the lenses required. If convex are accepted, hypermetropic; 
if concave are required, myopic. 



A child in good health shows under objective test by the static 
method 5 D. of hyperopia and with the Maddox rod test or a similar 
test esophoria of 8 degrees, what correction would you prescribe? 

The fact that hypermetropia and esophoria exist together 
shows that the functions of accommodation and convergence 
maintain their proper relation to each other. Under such cir- 
cumstances we are usually justified in giving the full correction 
for the refractive error. Whether or not we would prescribe exactly 
the + 5 D. would depend upon the age of the patient and the 
effect of the lenses on the acuteness of vision. If the child was 
over 6 and the acuteness of vision was markedlv diminished bv the 



188 State Board Examinations 

lenses, we might feel it advisable to reduce them, whereas younger 
children bear stronger lenses and possible blurring with little 
complaint. 

Explain the uses of the stenopaic slit and the pinhole disk and 
give the practical value of each. 

The stenopaic slit is intended to restrict the rays of light 
entering the eye to one meridian and in this way we can study 
the acuteness of vision and refraction of that one particular 
meridian. When placed at 90 degrees the vertical meridian comes 
under observation ; when at 180 degrees, the horizontal. Its 
special field is in the detection and correction of astigmatism. 

The stenopaic slit is placed in the Trial Frame and rotated 
through the different meridians, w^hile the patient looks at the 
letters on the test card. In this way each meridian for the moment 
performs the act of vision, and if there is any difference in vision 
in the various meridians, getting better in one w^ay and getting 
worse in the other, astigmatism is shown to be present. For 
instance, when at 90 degrees vision may be 20/20 and at 180 
degrees, 20/30. The spherical lens placed in front of the slit will 
measure the refraction of each meridian, and the difference be- 
tween the two lenses will show the amount of astigmatism. 

The pinhole disk is made use of in cases of greatly impaired 
vision to determine if the trouble is due to an error of refraction 
or not. If the pinhole raises vision, then glasses w^ill be of bene- 
fit; but if it fails to cause any improvement then there is little 
use to try to prescribe glasses. This is a simple and practical 
method to determine if the impaired vision is due to disease or 
to refractive error. 



How may it he positively determined {subjectively) whether or 
not ametropia is present when vision is 20/20 without lenses? What 
may he the character of the ametropia? 

It is possible for vision to equal 20 20 in presence of hyper- 
metropia or slight hypermetropic astigmatism because the accom- 
modation is instinctively brought into action to overcome the 
deficient refraction and maintain vision at the normal standard. 
In such cases weak convex lenses may be accepted, but just as 



Practical Optometry 189 

often they are rejected; the proper way to positively determine 
the presence of hypermetropia, which may be latent, is by means 
of the fogging system. 

A patient 35 years of age having 3 diopters of hypermetropia 
asks for ''reading' glasses hut complains of constant migraine; give 
the correction that you would order and state, with reason, whe?i in 
your opinion the glasses should be worn. 

The fact that the patient suffers with constant headache 
would be sufficient reason for advising glasses to be worn con- 
stantly. We do not know what method was used to measure the 
3 D. of hypermetropia, nor what the acuteness of vision was 
naturally and with glasses. Perhaps the patient will not bear 
the full correction for constant wear, in which case we would order 
as strong as can be comfortably worn for that purpose, and the 
+ 3 D. as an additional pair for reading. As patient will probably 
object to two pairs of glasses, we would theoretically be justified 
in ordering the + 3 D. for constant wear and also for reading, but 
in practice we are sometimes compelled to modify as above. 



// there is a tendency of the eyes to cross, on which side will the 
light streak appear with the Maddox rod over the left eye? How should 
the prism he placed to bring the lights together? 

By ''tendency of the eyes to cross" we understand a tendency 
to convergent strabismus; this would result in homonymous 
diplopia, in which right image belongs to right eye and left image 
to left eye. 

With Maddox rod over left eye, the streak of light would 
appear to the left; this is corrected by prism placed base out, 
because prisms displace objects toward their apex. 



State (a) how the amplitude of accommodation is measured, 
(b) how convergence is measured. 

(a) The nearest point for which the eye can accommodate 
by the strongest effort of the ciliary muscle represents the ampli- 
tude of accommodation. As explained above, if this should 



190 State Board Examinations 

be 5 inches, then 8 D. is the measure of the ampHtude of accom- 
modation; if at 4 inches, then 10 D. of accommodation. 

(b) The degree of convergence is measured by the angle 
through which each eye must turn from paralleHsm of the visual 
lines in order to converge to the near point of fixation. The ampli- 
tude of convergence is the distance from the far point to the near 
point of convergence, and is expressed by the greatest number of 
meter angles which the eyes can exert, the meter angle being 
that angle through which the eyes must turn from parallelism 
to converge at a point one meter away. At 20 inches {}4 meter) 
there would be 2 meter angles of convergence; at 10 {% meter) 
4 meter angles, and so on. In emmetropia the convergence and 
accommodation are equal; for instance, at 10 inches there are 
4 meter angles of convergence and 4 D. of accommodation. 

The convergence can also be measured by the strongest 
prisms, bases out, which the eyes are able to overcome and 
maintain single vision. 

Give reasons for and against full correction of myopia (a) in 
orthophoria, {b) in heterophoria. 

(a) When orthoporia exists in connection with myopia, the 
indications would be against the full correction of the refractive 
error, all other things being equal, because the normal relation 
between accommodation and convergence has not been main- 
tained. 

(b) If esophoria exists in connection with myopia, the indi- 
cations are still stronger against the full correction of the refrac- 
tive error, because the relations between it and the muscle balance 
are just the reverse of what they should be. 

When exophoria exists in connection with myopia, full cor- 
rection of the latter is indicated, because the relations are normal, 
and the concave lens also corrects the tendency to outward devia- 
tion. 

If an addition to the prescription for distance glasses must be 
made for presbyopia of -{- 3.50 D. or more, what error in the prescrip- 
tion for distance should be suspected? 

For the correction of simple and uncomplicated presbyopia, 
glasses are seldom required stronger than + 3 D. Therefore, in 



Practical Optometry 191 

the case mentioned where + 3.50 D. or more must be added for 
reading, it is fair to assume that the hypermetropia has not been 
fully corrected and that the distance glasses are not as strong as 
they should be. 



State {not describe) the various methods you would employ in 
making a thorough examination of the eyes. Write a hypothetic 
prescription involving sphero- cylinder prisms in bifocal form and 
give a complete case record of the same. 

i\cuteness of vision. 

Amplitude of accommodation. 

Ophthalmometer. 

Ophthalmoscope. 

Retinoscope. 

Trial case examination. 

Muscle tests. 

O. D. = 20^'60 - 1 D. sph. - 50 D. cyl. axis 180° = 20 20. 

O. S. = 20/40 - 50 D. sph. - 50 D. cyl. axis 180° = 20/20. 

Reads .50 D. type, 10" to 30". 

Ophthalmometer shows .50 D. excess in vertical meridian. 

Ophthalmoscope, normal fundus. 

Retinoscope, against movement. 

R. E., — 1 horizontally, — 1.50 vertically. 

L. E., — .50 horizontally, — 1 vertically. 

Maddox rod = 4° exophoria. 

Prescription given as follows: 

O. D., - 1 D. sph. - .50 D. cyl. axis 180° C Pr. 1° b. in. 

O. S., - .50 D. sph. - .50 D. cyl. axis 180° C Pr. 1° b. in. 

+ 1.50 D. segments added. 



After fogging the vision with plus lenses how do you proceed and 
when do you stop? 

Proceed to neutralize the excessive convexity and reduce the 
fogging by means of concave lenses placed in front of the fogging 
lens and gradually increased until a vision of 20/20 is reached. 
We must stop here or the purpose of the fogging system would be 



192 State Board Examinations 

defeated. Stronger concaves would be accepted if used but they 
must not be tried. 

About how much difference can he made in glasses when the 
eyes are not alike in focus? 

If a rule must be given we would say from 1.50 D. to 2 D., 
but there is a wide difference of opinion on this point. The writer 
has had cases where a difference of 4 D. and 5 D. has been made. 
After all it is a matter that must be left to experience in each 
individual case. The effort should always be made to give each 
eye its own proper correction. If such a difference proves uncom- 
fortable then the glass of the poorer eye must be changed but only 
so much as to make it bearable. 



Do you prescribe the weakest or strongest lens with which a 
patient can see 20/20 in hypermetropia and why? 

We give preference to the stronger glass in order to relieve 
the unnatural tax upon the accommodation and because the 
strongest convex lens is the measure of the manifest hyperme- 
tropia only and there is usually some latent error back of it which 
is not discovered by any lens. 



In myopia? Why? 

In myopia we give preference to the weakest glass in order 
to lessen the tax upon the accommodation and because there is 
usually some spasm of the ciliary muscle, which makes the myopia 
appear greater than it actually is and thus leads to the choice of 
a concave glass that is stronger than enough to correct the real 
amount of myopia. 

What is presbyopia and how corrected? 

Presbyopia is that condition of vision where on account of a 
failure of accommodation, due to a loss of contractility of the 
ciliary muscle and of elasticity of the crystalline lens, the near 
point recedes to an inconvenient distance and reading is impaired 
or made impossible. 



Practical Optometry 193 

It is corrected by means of a convex lens, which restores the 
receded near point, assists the accommodation and places in front 
of the eye the necessary convexity which has been lost within it. 



How do you test the external and internal muscles? 

By prisms, with apex over the muscle it is desired to test. 



How many degrees of prism, base out, should a person overcome? 
20° to 30°. 



How many degrees of prism should a person overcome base in? 
Up or down? 

6° to 8° base in. 

2° to 3° up or down. 



// the eyes turn in, how do you place the base of a prism? 
Out, the rule being base opposite to deviation. 



What is the object of fogging with strong plus lenses? 

To encourage relaxation of the ciliary muscle and reduce 
spasm of the accommodation. 



How do you test for astigmatism? 

Objectively by the ophthalmometer and the retinoscope and 
subjectively by the radiating lines and cylindrical lenses. 



How would you use the stenopaic slit in finding the error of 
refraction in an eye which has astigmatism? 

The stenopaic slit is placed in position before the eye of 
the patient, who is asked to look at the test letters. It is then 
rotated to that place where the letters are seen the plainest, which 
will be one of the principal meridians and that of best vision. 



194 State Board Examinations 

The meridian at right angles will be the other principal 
meridian and that of poorest vision. The refraction of each 
meridian is measured by spherical lenses. 

If the vision in the first meridian is normal, this meridian 
may be emmetropic or hypermetropic. The vision in the second 
meridian is necessarily below normal, and may be hypermetropic 
or myopic. 

Convex lenses are tried in the first meridian; if rejected we 
assume this meridian is normal and the case one of simple astig- 
matism. If accepted, this meridian is hypermetropic and the 
astigmatism is either compound or mixed. 

The second meridian is measured by convex and concave 
lenses in the usual way. If convex lenses (of different strengths) 
are accepted in both meridians, the astigmatism is compound; if 
convex lenses in one meridian and concave in the other, mixed. 

It must be remembered that the axis of the correcting cylinder 
is to be placed at right angles to the meridian that is being 
measured. For instance, if + .50 D. is accepted in the vertical 
meridian, the correcting lens would be + .50 D. cyl. axis 180°. 

If + 1 Di is accepted in the horizontal meridian, the cor- 
recting lens would be + 1 D. cyl. axis 90°. 

And the prescription would be + .50 D. sph. O + .50 D. 
cyl. axis 90°. 



What is understood hy the so-called ''cover'' and ''fixation' 
tests for muscular weaknesses? 

In any case of muscular w^eakness we assume the good eye 
to be the fixing eye. The patient is asked to look at an object 
some distance, a light being the best. A card is then placed over 
the fixing eye, thus excluding it from vision, when the patient 
will be compelled to use his other eye to see the light. If this 
causes any movement in either eye there is evidence of muscular 
insufficiency. Both eyes should be covered in turn, and if there 
is no movement of either eye after covering or uncovering, both 
eyes would seem to fix, thus contraindicating insufficiency. 

If either eye turns inward w^hen its fellow^ is covered, it 
must previously have been deviating outwards; and if the excur- 
sion should be outward, the previous deviation must have been 
inward. 



{ 



Practical Optometry 195 

In the fixation test both eyes look at an object, the distance 
of which is quickly changed, while the corresponding movements 
of the two eyes are closely watched, according to which we can 
detect the presence of insufihciency. 



What test may be applied to determine whether or not an 
anisometrope will accept his full correction for widely different 
states of refraction? 

There is no definite test at our command by which we can 
determine this point. It is a matter that really can be decided 
only by the experience of the patient. The glasses should be 
tried for ten or fifteen minutes in the ofifice, and if there is no 
decided objection they may be given tentatively for a few days' 
or a few weeks' wear. If there is a slight discomfort at first, 
this may pass away after a few days wearing of the glasses. 
At any rate a reasonable efTort must be made to have the eyes 
bear the full correction for each eye. 



What do you consider the most satisfactory way of correcting 
high myopia; full correction or partial correction? 

This depends upon a great many circumstances, such as age, 
amplitude of accommodation, whether glasses had been pre- 
viously worn, and if so, the proportion of full correction and the 
character of the work for which the eyes are to be used, as well 
as the degree of defect and the amount of vision obtainable. 

When patient first puts on glasses it is best to start with an 
undercorrection, which may be increased from time to time until 
the full correction is reached for distance. But for close use 
glasses 2 D. to 3 D. weaker should be used. 

It is understood that the stronger the concave lenses the 
greater the tax upon the accommodation, and as this function is 
less developed in myopia, caution must be observed not to 
increase this tax to the point of causing asthenopia. On the 
other hand, it may be possible to contribute to the ciliary in- 
sufihciency by weak lenses, or to develop the strength of the 
accommodation by lenses that are purposely made stronger for 
that purpose. 



196 State Board Examinations 

Name and describe the two principal divisions of hyperme- 
tropia. 

Hypermetropia as met with may be divided into manifest 
and latent. Manifest is that which is not concealed by the ciliary 
muscle/ and is determined by the strongest glass the patient can 
be induced to accept for distant vision. Latent hypermetropia is 
that which is neutralized by the power of accommodation, and 
usually in its entirety can be revealed only by the action of a 
cycloplegic. 

The manifest error may vary from time to time according 
to the tonicity of the ciliary muscle. In youth the manifest is 
proportionately small and the latent correspondingly large. With 
the advance of years the manifest increases at the expense of the 
latent, until finally the defect is all manifest. This condition is 
then known as absolute hypermetropia, when vision becomes 
impaired and convex lenses are a necessity. 



Explain the states of vision that should govern the decision as 
to whether or not lenses should be worn constantly. 

Theoretically, all errors of refraction call for constant 
wearing of glasses, but this is particularly true of hypermetropia 
and hypermetropic astigmatism; and yet even here in slight 
degrees of defect in the presence of a vigorous accommodation 
there may be exceptions to this rule. In mild cases of myopia 
where the distant vision is not inconveniently impaired, the 
constant wearing of glasses may be considered as much a luxury 
as a necessity. 

Independent of the condition of vision, and even when 
vision is normal, there are many cases that require the constant 
wearing of glasses to relieve the strain. 



Describe the methods of measuring the amplitudes of accom- 
modation and convergence. 

In order to measure the amplitude of accommodation the 
patient is given a card of very fine print and the very closest 
point at which it is possible to read this is measured in inches; 



Practical Optometry 197 

this is transposed into diopters, which will represent the amplitude 
of accommodation. 

To measure the amplitude of convergence a pin may be 
used and the closest possible point at which it is seen singly will 
be noted, and this corrected into meter angles. 



What error of refraction is often indicated on the surface of 
the cornea? 

Astigmatism. 



What is the condition called if the right eye deviates upward? 

Right hyperphoria, or if there is actual and noticeable 
deviation, right hypertropia. 



Name four methods of detecting hyperphoria. 

Maddox rod, Maddox double prism, phorometer and prism 



test. 



What advice would you give to a patient when he first wears 
glasses? 

This depends somewhat on the character and degree of the 
refractive error. In hypermetropia and astigmatism we would 
tell him the glasses may seem strange at first and that it will 
take him a little time to get accustomed to them. We will 
warn him that the ground may seem to slant as if he were walking 
up hill and there may be some distortion of objects, but that if 
he will persist in wearing the glasses for several days these un- 
pleasantnesses will most likely gradually pass away. In many 
of these cases the natural acuteness of vision is normal, hence in 
order that a patient may not be disappointed it is well to tell 
him that the glasses are not intended to make him see better, 
but to relieve the strain and discomfort from which he suffers. 



In making a rapid preliminary examination what would 
cause you to suspect high myopia? 



198 State Board Examinations 

A great impairment of distant vision, so that even the largest 
letters on the test card could not be named; a tendency to bring 
reading and small objects very close to the eyes, and a squinting 
or half closing of the lids; a dull, slow movement against the 
mirror with the plane retinoscope; a rejection of convex lenses 
and a marked improvement in vision by concave lenses. 



When will a hyperope accept minus glasses for distance? 

When there is spasm of accommodation, as there usually is 
in hypermetropia. If the accommodation was passive rays of 
light would strike the retina before they could come to a focus, 
resulting in a blurred image; but as soon as the eyes are opened 
the accommodation at once comes into action to increase the re- 
fractive power of the hypermetropic eye and bring the rays to a 
focus on the retina. It sometimes happens that this action of 
the accommodation is overdone; that is, more effort is exerted 
than is just necessary to neutralize the hypermetropia, and then 
a condition of false myopia is produced when concave lenses 
would be accepted. It goes without saying that even if accepted, 
concave lenses should not be prescribed in hypermetropia. 



What is astigmatism? 

That condition of the eye where its refraction is similar to 
a sphero-cylindrical lens, where there are two meridians of 
greatest and least curvature, the difference between which 
represents the amount of astigmatism, and as a result the rays 
cannot be focussed at a single point, but there is a focus for each 
meridian. 

In the practice of optometry what is sought to he done hy the 
use of lenses? 

To correct ametropia and heterophoria. To add to the 
refractive power where deficient, as in hypermetropia; to lessen 
the refractive power where excessive, as in myopia, and to 
equalize the refraction in all the meridians of the eye, w^hen one 
or more are in error, as in astigmatism. In other words, to make 
the eyes emmetropic. In heterophoria to assist the deficient 



1 



Practical Optometry 199 

muscles by prisms in position of relief and thus restore the 
proper equilibrium; also to develop these muscles by prisms in 
position of exercise. To neutralize any deficiency in accom- 
modation, as in presbyopia. In fine, to improve vision or make 
it more comfortable bv remo\'ino^ strain. 



In testing with the stenopaic slit it is found that the right eye 
requires + 1.75 in the vertical meridian aiid — .75 in the horizontal. 
What is the prescription? 

+ 1.75 D. cyl. ax. 180° C - .75 D. cyl. ax. 90° 
This cross cylinder may be transposed into either of the following 
sphero-cylinders : 

+ 1.75 D. sph. C - 2.50 D. cyl. axis 90° 
- .75 D. sph. C + 2.50 D. cyl. axis 180° 



What rule is to he observed in the giving of minus lenses in 
myopia? 

When a concave lens of the proper strength to exactly 
neutralize the myopia is placed before such an eye. parallel rays 
are diverged just enough to throw the focus from in front of the 
retina back on to the retina. If the concave lens is any stronger 
than necessary the rays are made more divergent, which would 
tend to cause them to focus behind the retina; but the accom- 
modation instinctively comes into action to bring the focus of 
these divergent rays forward to the retina. This led to the rule 
that in myopia the weakest glasses that afford satisfactory vision 
should be prescribed in order to avoid a tax upon the accom- 
modation. 

It may be argued that a hypermetrope sees better with a 
glass that is not a full correction and which allows him to use 
some of his accommodation, and hence there is no harm in an 
over-correction of myopia where vision is made perfectly clear 
by a slight use of the accommodation. But it must be remem- 
bered that the former is accustomed to accommodating and his 
ciliary muscle is large and well developed, while in the latter 
case there is but little need for accommodation and the ciliary 
muscle is small and weak. Of course, if glasses are placed in 



200 State Board Examinations 

youth on a patient who is vigorous the ciliary muscle is likely 
to develop the normal power, but if glasses are first worn later in 
Hfe, or in any case as age creeps on, the glasses should be kept 
weak, or reduced for close use. 



What is indicated when the pinhole disk alone produces marked 
improvement of vision ? 

That the impairment of vision is due to an error of refraction 
and that it can be corrected by lenses. 



The patient is directed to look at a light 20 feet distant and 
is given diplopia with a prism, base down, before the right eye; what 
defect and amount of error are present when the upper light is 
located to the left and a prism of 5° before each eye is required to 
bring both lights to the median line? What is the position of the 5° 
prisms? 

An artificial vertical diplopia is produced by the prism base 
down before right eye. The upper light will belong to this eye, 
and if it is located to the left a condition of crossed diplopia is 
shown to be present, due to an exophoria. The amount of the 
imbalance is 10° and the prisms are placed bases in. 



A hyper ope of 2.5 D. sees with his right eye the Maddox-rod 
line 6 cm. higher than the object light; write the prescription for 
distance lenses, eye size, so that the least cost will be incurred. 

Inasmuch as the image formed in the right eye is the highest, 
the condition is one of left hyperphoria, which can be corrected 
by prism base down left eye or base up right eye. And as it is 
desired to write a prescription incurring the least cost, it is 
necessary to obtain this prismatic effect by a decentering of the 
lenses. 

No mention is made of the distance at which the test is 
made, but we will assume the usual distance of 20 feet or 6 meters. 
The deviation for one prism diopter is just 1 cm. for each meter 
of distance; hence for 6 meters there is 6 cm. of displacement. 
Therefore, in this instance where there is 6 cm. of deviation, 
there is indicated 1 p. d. of insufiiciency. 



Practical Optometry 201 

The rule for decentration is as many degrees of prism power 
as there are diopters of refractive power for every decentration 
of 10 mm. In this case the combined value of the two lenses is 
5 D., which would afford 5° of prismatic power if decentered 
10 mm., or 1° prismatic power if decentered 2 mm. As this is 
the amount desired, it can be obtained by decentering the right 
lens, 2 mm. up, or the left lens 2 mm. down, or dividing the 
decentration between the two eyes, right eye 1 mm. up and left 
eye 1 mm. down. 

What are the conditions of vision which make the use of distance 
glasses imperative? 

When the acuteness of vision is so greatly impaired by 
marked degrees of myopia, hypermetropia or astigmatism as to 
interfere with the ordinary occupations of life, or to make walking 
in a crowded street difficult or dangerous. 



When may distance glasses he dispensed with? 

Principally in hypermetropia in the slighter degrees and 
especially in young people where the accommodation is active 
and vigorous, and no symptoms of asthenopia are in evidence. 



A person aged 40 can just see distant objects clearly with a 
minus 1.50 D. lens. What glasses, if any, would be required for 
reading at 14 inches? What reading glass wotdd he required at 
double his age? 

We will assume that the — 1.50 D. lens represents the amount 
of myopia present. 

The average amplitude of accommodation at 40 years of age 
is 4.50 D., which in this case is increased by the myopia to 6 D., 
of which he is able to use comfortably from one-half to two-thirds, 
or 3 D. to 4 D. At 14 inches 2.75 D. of accommodation is neces- 
sary, so that it is evident there is still a surplus of accommodation 
and no glasses will be necessary for reading. 

At 80 years of age, when the accommodation is entirely lost, 
he would theoretically need + 1-25 D. lenses, which added to the 
1.50 D. of myopia would give the 2.75 D. necessary for reading 



202 State Board Examinations 

at 14 inches. But practically on account of the senile diminution 
of refractive power it is likely the glasses will have to be stronger 
than 1.25 D. 



In cases of esophoria where is the base of the prism to he placed, 
in or out, and how should strength or number of prism be determined 
for exercising the same, and the proper way of exercising? 

For the correction of esophoria the base of the prism is placed 
out. 

If the esophoria was supposed to be due to weakness of the 
external recti and it was desired to exercise the same, the apex 
of the prism must be placed over the muscle to be exercised, which 
in this case would be base in. 

First it must be determined how much these muscles are 
able to overcome. Normally these muscles have a strength of 6° 
to 8°. If they were weak this might be reduced to 1° or 2°, but 
this must be determined as a starting point, and then the strength 
of the prisms gradually increased until the normal standard is 
reached. 

A certain patient is astigmatic, presbyopic and hypermetropic. 
In what order are these defects tested for? 

We examine for errors of refraction first and of accommoda- 
tion next. Of the errors of refraction we commence with the 
spherical errors, hence the order would be: Hypermetropia, astig- 
matism and presbyopia. 

// a 6° prism is placed in front of an eye with both eyes in use, 
in what direction does the base of the prism have to be in order that 
there might be diplopia? 

If placed base out, it would be quickly overcome by the 
internal recti. If base in, there is some question whether it could 
be overcome by the external recti, and there might be diplopia. 
But if placed vertical there would surely be diplopia. 



An eye is hypermetropic 4 D., of which 1 P.is latent. What will 
be his far point with a -\- 6 D. lens? 



Practical Optometry 203 

If the latent hypermetropia refused to accept correction there 
would be 3 D. of manifest hypermetropia. Then the 6 D. lens 
would represent an overcorrection of 3 D., or an artificial myopia 
of this am.ount. which would be represented by a far point of 13 
inches. 



What proportionate amount of ametropia should he corrected in 
a person 20 year of age, and hoi^' much should be corrected in a person 
50 years of age? 

In a person 20 years of age we might correct one-half to 
two-thirds of the ametropia, or perhaps it would be safer to say, 
the manifest defect. In a person 50 years where the accommoda- 
tion for distance need scarcely be reckoned with, the full correction 
may be given. 

If an additioyi to a prescription for distance glasses must he 
made for presbyopia of + 3.50 D. or more, -uhat error i^i the pre- 
scription for distance should be estimated? 

If the distance glasses were convex we would suspect that 
the hypermetropia had not been fully corrected, because the 
presbyopic addition is scarcely ever more than 3 D. 



What is the Maddox rod, and for zvhat is it used? 

It is a strong cylinder of glass, either white or red, which 
elongates a point of light into a long, narrow streak of light at 
right angles to its axis, and is used in the detection of the various 
forms of heterophoria. 



Can the subjective test be made at less than 20 feet; that is 
where 20 feet is not available? 

Yes, if allowance is made for the increased divergence of the 
rays from the shorter distances. For instance, if the test is made 
at a distance of 10 feet the rays would have a divergence of 120 
inches, which is equal to .?>?> D., which must be subtracted from 
the convex correction and added to the concave. 



204 State Board Examinations 

Some operators prefer to use a mirror, which in a 10-foot 
room gives the effect of 20 feet. 



Where should the base of the prism be placed in measuring 
powers of adduction, abduction, supraduction and infraduction? 

The apex of the prism is placed over the muscle to be tested ; 
hence adduction is measured by prism base out, abduction base in, 
supraduction base down, and infraduction base up. 



Which test do you think the most accurate for muscle imbalance? 

There may be some difference of opinion on this point, but 
for all practical purposes we should say the Maddox rod test. 



In cases of anisometropia what is the greatest amount of dif- 
ference in lenses which can usually be worn for comfortable binocular 
vision, and why? 

This is an open question, as the difference varies in each case. 
Some authorities would fix the limit at 2 D., but we have knowl- 
edge of cases where the difference was as great as 5 D. This is a 
point that cannot be determined by any rule, but only by actual 
trial. 

Is the test with the perimeter subjective or objective, and why? 

This is a subjective test, because the information to be 
gained is dependent upon the answers of the patient. 



A myope of 5 D. has an amplitude of accommodation of 2 D, 
What lenses would you prescribe for reading music at 50 cm. so as 
to permit the use of one-quarter of his power of accommodation? 

The full amplitude of accommodation is 2 D., but he is to 
be permitted to use only .50 D. In order to see at 50 cm. without 
any accommodation a+ 2 D. lens is required, but as this patient 
is allowed to use .50 D. of his accommodation then only + 1.50 
D. is necessary, which added to his distance glasses would reduce 
them to - 3.50 D. 



Practical Optometry 205 

How should preshyope he tested for muscle balance at fourteen 
inches? 

The usual method is by means of the dot and line test and a 
vertical prism. Of course, the necessary convex lenses must be 
worn and care be taken to see that they are properly centered for 
the fourteen-inch distance, as otherwise the lenses themselves will 
introduce a prismatic effect. 

The Maddox rod may also be used with a small point of light 
at the indicated distance of fourteen inches. 



Under what conditions would you consider it unwise to advise 
the use of distance glasses, even though ametropia were present? 

If any disease of the eye calling for the attention of a medical 
man was known to be present, or even strongly suspected, it 
would not be wise for an optometrist to prescribe glasses, even 
though an error of refraction was also discovered. 



Write a hypothetic prescription involving prisms, spheric and 
cylindric lenses. 

+ 1.50 D. sph. C + .50 D. cyl. axis 90° C prism 2° base in. 



A -]- 4 D. sphere is placed before an eye that is 2 D. hyperme- 
tropic, and an object twenty inches away is looked at. What will 
be the character of the image on the retina? 

In order to see an object at a distance of 20 inches the 
emmetrope will use 2 D. of accommodation; the hypermetrope, 
in addition, must use sufficient accommodation to overcome his 
error; therefore, in the case mentioned, 4 D. of accommodation 
will be required at 20 inches. If now a + 4 D. lens is placed 
before the eye the result will be a distinct image upon the retina 
without effort of accommodation. 



When the targets of an ophthalmometer overlap at an angle of 
120° and separate at 30°, what will be the axis of the concave cylinder? 

This would indicate that the greatest refraction was located 
in the 120th meridian and the least in the 30th meridian; there- 
fore, the axis of a concave cylinder would coincide with the latter. 



206 State Board Examinations 

Upon what general principles are the various tests for muscle 
imbalance based, and what is a main objection to these tests? 

There are two general principles on which the various 
muscle tests are based : 

1. Those which displace the image in one or both eyes from 
the macula and thus produce an artificial diplopia. 

2. Those in which the image formed in one eye is changed 
in color, size or shape, which dissimilarity in the two retinal 
images causes diplopia. 

In both instances binocular vision is destroyed, and a weak- 
ness in any of the muscles is supposed to manifest itself. 

The objection to the first class of tests is the displacement 
from the macula, but in both of them we have no means of know- 
ing how much innervation is sent to the several muscles and how 
much of the heterophoria may thus be made latent. 



If a presbyope wearing a convex lens moves it farther from his 
eye what is the change in dioptric power? 

The effect of moving a convex lens farther from the eye is 
to increase its dioptric power. 



If a + .50 D. sphere gives best vision on the horizontal lines 
of the astigmatic fan, and a -\- 1 D. sphere best vision on the vertical 
lines, what is the prescription for the correcting lens written in 
three different ways? 

+ .50 cyl. axis 180° C + 1 cyl. axis 90° 
+ .50 D. sph. C + .50 D. cyl. axis 90° 
+ 1 D. sph. C - .50 D. cyl. axis 180° 



In trying minus lenses on a myope what principal rule would 
you constantly keep in mind? What rule in cases of hyper metro pia, 
using plus lenses? 

In myopia the weakest concave, to impose as little tax upon 
the accommodation as possible. 

In hypermetropia the strongest convex, to assist the accom- 
modation as much as possible. 



Practical Optometry 207 

The distance test shows that a certain patient is 1.25 D. hyper- 
metropic, hut the dynamic test with the cross cylinder at 20 inches 
shows that a + .25 sphere cause neutrality. What glasses would 
you prescribe? 

Not stronger than + .25 D., at least to start with. 



What are ''cover'' tests and what is their object? 

A screen or card is placed before one eye, which is watched 
to discover if it makes any movement when thus covered. If 
not orthophoria is indicated. If it moves outward there is a 
presumption of esophoria; if it moves inward a presumption of 
exophoria. 



On what principle does the Maddox rod test act? 

That it causes a retinal image so dissimilar in size, shape 
and appearance from the other that the natural instinct to fuse 
them into one is for the time being destroyed, and in this way a 
deficiency in any of the muscles becomes manifest. 



Under what circumstances would you recommend the constant 
wearing of glasses? 

When distant vision is impaired, when headache and 
symptoms of asthenopia are present, in high errors of refraction 
and usually in all cases of astigmatism. 



What acuity of vision would you expect to find in a young 
case of mixed astigmatism, in which the corrected cylinder would be 
50 D. or less, with the ride? Against the rule? 

In a case of mixed astigmatism the accommodation is apt 
to be brought into play to overcome the hypermetropic meridian, 
and in so doing it makes the myopic meridian more myopic. 
When the astigmatism is with the rule it is the vertical meridian 
that is myopic, which is the meridian that focuses horizontal 



208 State Board Examinations 

lines. But as the visibility of letters largely depends upon their 
vertical lines, under these circumstances vision would be little 
if any impaired. 

In astigmatism against the rule the horizontal meridian is 
the myopic one, and as this is the meridian that focuses the 
vertical lines vision is more likely to be impaired. 



Under what circumstances should hypermetropes be given full 
correction? 

When patient is past middle age, distant vision impaired 
and the hypermetropia practically all manifest. 

When a high degree of esophoria or convergent strabismus 
is associated with the refractive error. 

When a partial correction fails to afford relief. 



What is the effect on the ciliary muscles of giving a full correc- 
tion in the case of hypermetropes? 

The ciliary muscles are relieved of all effort in distant vision, 
and are called into use only for the accommodation necessary 
in near vision, just as is the case in emmetropia. 



Where is the base of the prism placed when the eye turns in? 
That is, for constant wear. And where placed to exercise the 
muscles? 

Base out for constant wear and base in for exercise. 



How can we with a stenopaic disk find an error of a vertical 
meridian in a case of astigmatism which requires for correction 
a + 50 D. cylinder axis 180°? 

The stenopaic disk is placed over the vertical meridian of 
the eye and the amount of hypermetropia in this meridian 
measured by the acceptance of convex lenses, or, better still, 
by the fogging method. 

In a muscle test with a single prism over the right eye base 
down the upper image is seen on the left. What is the nature of the 
heterophoria and how can it be measured? 



Practical Optometry 209 

The upper image is that of the right eye, and being seen on 
the left eye indicates exophoria. It can be measured by strength 
of prism base in that it brings the two hghts in the same vertical 
plane. 

How can the near point of convergence be found objectively? 
Subjectively? 

By the near point of convergence is understood the closest 
point for which the eyes can converge. In order to measure it a 
small fixation object is held in front of the eyes and gradually 
brought closer and closer. 

In the objective test the eyes are sharply watched, and as 
soon as one of them ceases to converge or commences to diverge 
the near point of convergence has been reached. 

In the subjective test the object is approached until it is 
seen double by the patient. 



A myope of 4 D. has an amplitude of accommodation of 3 D. 
He is to be fitted with glasses to read at 16 inches, but must use one- 
half of his accommodation. What must be the power of his glasses? 

The positive refractive power of such an eye would be equal 
to the sum of the amplitude of accommodation and the myopia, 
or 7 D. In order to read at 16 inches 2.50 D. of accommodation 
is necessary, but if he must use only one-half of his accom- 
modation then he should possess 5 D. In order to bring about 
this result the power of the glass required would be — 2 D., 
which reduces the 7 D. of accommodation to the desired amount. 



When plus lenses are decentered in, what extrinsic muscles are 
assumed to be at fault? 

The effect produced being that of prisms bases in, the 
internal recti would be assumed to be weak. 



A patient giving his age as thirty -five years requires + .50 D. 
for distance and -\- 2 D. for reading. What would you suspect? 



210 State Board Examinations 

Would suspect that the + .50 D. did not represent the full 
correction for distance, but that the amount of hypermetropia 
was more nearly equal to 2 D. and that the proper tests had 
not been made to uncover the latent error. 



In tiiv given cases, each patient's age being thirty years, neither 
having worn glasses before, normal vision is obtained with spherical 
correction. One is hypermetropic 2 D. and has 2° of exophoria; 
the other is myopic 2 D. and has 2° of esophoria. Write prescrip- 
tions for both. 

When hypermetropia occurs in connection with exophoria, 
and myopia in connection with esophoria, there is a disturbance 
of the normal relation that should exist between the accommoda- 
tion and the convergence. 

In the first case we would be inclined to undercorrect 
the hypermetropia because convex lenses tend to aggravate 
exophoria; or we w^ould combine a prism base in to correct the 
exophoria, not exceeding one-half the amoimt of insufficiency. 

In the second case we would also think of undercorrection, 
because a concave lens tends to increase an esophoria ; or we would 
combine a prism base out with the concave lens, not attempting 
to correct more than one-half the heterophoria. 



What course should be followed in prescribing for anisometropes 
whose eyes differ widely in refraction? 

Correct the best eye fully and the other eye approximately. 
In other w^ords, take care of the good eye and do what you can 
for the poorer eye at the time, with the thought of gradually 
increasing the correction of the latter, as the comfort of the 
eyes wdll allow. The visual acuity of each eye, as well as the 
muscle balance, must be taken into account. It is impossible 
to give general directions that will apply to all cases, but each 
case must be managed on its own merits. 



{ 



Give reasons why it is sometimes not advisable to prescribe 
distance glasses. 



Practical Optometry 2 1 1 

When the acuteness of vision is not impaired, when the 
refractive error is small, when there is no headache or photo- 
phobia or evidence of eyestrain in general vision, and when 
glasses cause so much fogging of distant vision as to be unbear- 
able. 

Under n'hat conditions should the constant wearing of glasses 
be insisted on? 

In case of strabismus due to refractive error, in almost all 
cases of astigmatism, when headache is present due to eyestrain, 
and w^hen vision is very greatly improved by the glasses. 



Under what circumstances might a concave lens be prescribed 
for a hypermetrope? 

The optical student is so much cautioned about the danger 
of giving concave lenses to a hypermetrope that when he gets 
into practice he would not want to be accused of such ignorance 
as this question would seemingly display. And while this 
teaching is correct, there are cases of hypermetropia with in- 
sufhcient convergence, in which it might be allowable to prescribe 
concave lenses for temporary use for their indirect effect of 
stimulating the conv'ergence. 



Give reasons for a?id against full correction of myopia and 
hypermetropia in (a) orthophoria and (b) heterophoria. 

{a) In orthophoria, we give the strongest convex lens in 
hypermetropia and the weakest concave lens in myopia, in order 
to assist the accommodation or lessen the strain upon it. 

{b) In heterophoria, the above rule must be modified in 
the light of the effect of the lenses upon the accommodation and 
convergence. 

When hypermetropia has esophoria associated with it, the 
full correction is given; but when exophoria is present, it is best 
to under-correct, because convex lenses aggravate exophoria. 

In the case of myopia with exophoria, the full correction 
may be given ; but when esophoria is present, the concave lenses 



212 State Board Examinations 

should not be too strong, because concave lenses aggravate 
esophoria. 

What is the usual method of employing the stenopaic slit in 
testing the eye, and why is it not in common use? 

While the patient looks at the card of test letters the slit 
is rotated to the meridian of best vision, in which position trial 
lenses are used to determine if this meridian is emmetropic, 
hypermetropic or myopic. The slit is then rotated to the meridian 
at right angles where test lenses are used to determine if it is 
hypermetropic or myopic. Having estimated in this way the 
refractive condition of the two principal meridians, it is easy 
to formulate the cylinder or sphero-cylinder that is required to 
correct. 

It is scarcely reliable because the accommodation is likely 
to come into play and may be used more in one meridian than 
the other, thus impairing the accuracy of the test. 



When may myopes he given the strongest possible lenses to 
produce normal vision? 

The standard advice to prescribe the weakest concave 
lenses in myopia may be modified in young people when exo- 
phoria is present. 

Of two equally far sighted persons, one has the habit of wearing 
his spectacles low down on his nose, the other wears them close to 
his eyes. Which should have the stronger spectacles, the object 
being held at the same distance from the eye by both persons? Give 
the reason for your answer? 

It is a well-known fact that convex lenses increase in power 
as they are pushed farther from the eye. In these cases we are 
led to infer that they are both under-corrected. In the first 
case the person has learned that he can get the extra power that 
is needed by wearing his spectacles low down on his nose. 

The second person either has not learned this trick or he 
prefers to wear his glasses in their proper place close up to the 
eyes, and therefore he is the one that should have stronger 
spectacles. 



Practical Optometry 213 

A patient with a pupillary distance of 59 mm. wears -\- 4 D. 
lenses for distance in a spectacle frame with a pupillary width of 
61 mm. What is the effect of the lens and what is the amount? 

Under these conditions patient would not look through 
optical centers of lenses and therefore a prismatic effect would 
be produced; and as the spectacle frames are too wide for the 
pupillary distance of the patient, and as the lenses are convex, 
the prismatic effect would be bases out. 

Figuring on the basis of 1° of prism for every 1 D. decentered 
10 mm., and as the amount of decentration in this case is 2 mm., 
the amount of prismatic power developed would be 4/5° for each 
lens. 

Patient aged forty-five has worn -\- 1 D. cyl. axis 180° with 
comfort for years previously, but now complains that reading 
in evenings is uncomfortable. What would be your treatment of the 
case? 

This patient has now reached the presbyopic period of 
life, and the cylinders worn for the correction of his refractive 
error will no longer sufffce for close use, but must be supplemented 
by the addition of convex spheres for the correction of the 
presbyopia, which at this age would probably be about + 1 D. 

I would re-examine his eyes and ascertain just what is 
needed for distance and for reading, and give him the choice 
between bifocals and two pairs of glasses. If separate pairs of 
glasses are ordered for close use, it is more than probable the 
cylinders may be reduced somewhat, as is usually the case with 
convex cylinders having a horizontal axis, because of the fact 
that when looking down through convex lenses, as in reading, 
there is produced the added effect of a cylinder with a horizontal 
axis, thus allowing the cylindrical surface of the lens to be reduced 
somewhat. 

In the case of astigmatic presbyopes is it always necessary to 
retain the cylinder in the reading glasses? And what is the cause 
of the difficulty experienced by presbyopes in getting accustomed to 
wearing cylinders? 

As a rule, if the astigmatism is of such a character as to 
cause symptoms of discomfort, it will be corrected earlier in life, 



214 State Board Examinations 

and then when the patient reaches the presbyopic age, it simply 
means the addition of a convex sphere of such power as will 
afford comfortable reading at the proper distance. 

But it sometimes happens that persons who have passed 
their fortieth year will apply for glasses, with the history that 
they have never worn them, or felt the need of them, but of late 
they are beginning to experience some difficulty in close vision. 

The optometrist starts to make his examination in the usual 
way and soon finds the existence of astigmatism, perhaps .50 D. 
to 1 D. He remembers that he was taught that astigmatism is a 
frequent cause of eyestrain and headache, and if he has not had 
much practical experience he will proceed to correct it and 
perhaps get himself into trouble, because the patient returns 
and complains that the glasses are uncomfortable. 

Another examination will show the presence of the same 
amount of astigmatism as at the first, and perhaps the patient 
will be advised to persevere with the sphero-cylinders, but with 
probably the same result as before. 

This brings up the question as to why it is more difficult for 
older persons to become accustomed to cylinders than younger 
ones. Now, it must be borne in mind that vision is not so much 
a question of physical optics as of physiological optics, and that 
there is a psychological feature as well; in other words, there is 
not only the formation of an image on the retina, but also the 
interpretation of that image by the brain. It is not the eye 
that sees but really the brain. 

Astigmatism causes some distortion in the retinal image, 
but it is not noticeable to the patient, because it has always 
existed, and, in fact, the sense of vision was acquired and culti- 
vated with that kind of an image. For instance, a square object 
would not form a perfectly square image in an astigmatic e3^e, 
but would be elongated in one direction or the other, according 
to the character of the error. 

The person, however, knows by education and experience 
that the object is actually square and his brain learns to interpret 
the slightly distorted retinal image as that of a square object, 
and he is satisfied with his vision. 

When the proper correcting cylinders are placed before the 
eye, the image of a square object is also square; but it is different 
from what the brain has been accustomed to recognize as a 



Practical Optometry 215 

square object, and hence, even though the image is square, the 
object appears to be elongated. 

The greater number of the cyHnders that are worn are 
prescribed more for the reUef of asthenopia than for the improve- 
ment in vision; but even in the higher degrees of astigmatism 
where the acuteness of vision is much less than normal, the 
correcting cylinders, though they afford perfect vision, are 
oftentimes disappointing and unsatisfactory for the reasons 
mentioned above. The patient is expecting great things and 
when he finds that the results are not as good as before, he is 
apt to become disgusted. 

A myope of 6 D. possessing 3 D. of accommodation, wants a 
pair of glasses for reading at 16 inches. If it is desired that only 
one-half the power of accommodation he tised, what would be the 
strength of the glasses you woidd give? 

In determining the reading glasses in high myopia, the rule 
is as follows: Subtract the glass representing the desired reading 
point from the full measure of the myopia. In this case the glass 
representing the desired reading point of 16 inches is 2.50 D., 
which we subtract from the full amount of myopia which is 6 D., 
and the result is — 3.50 D. 

With these — 3.50 D. glasses there remains an uncorrected 
myopia of 2.50 D., which represents a far point of 16 inches, at 
which distance the person would be able to read without any 
effort of accommodation. 

But we are told it is desired to use one-half of his 3 D. of 
accommodation and, therefore, to bring this amount of accom- 
modation into action, a — 1.50 D. lens must be added to the 
— 3.50 D. lens, which means that the glasses to be prescribed to 
accomplish the purpose indicated must be — 5 D. 



Can a concave lens have conjugate foci? If so, under what 
conditions? 

When parallel rays of light enter a concave lens, they are 
made divergent to such ^n extent as to appear to come from 
the principal focal distance of the lens. If the power of the lens 
is — 4 D., the rays after refraction will apparently diverge from 
a point 10 inches in front of the lens. 



216 State Board Examinations 

But when the rays proceed from an object nearer than 
infinity, in other words when the rays that enter the lens are 
already divergent, they are made still more divergent by the 
action of the lens. The divergence produced by the lens is 
increased by the divergence caused by the nearness of the object, 
and, therefore, the rays will appear to diverge from a point 
closer than the principal focus, which point would be conjugate 
to the object. 




Fig. 32 



In this diagram it will be seen that the rays diverging from 
fi, which is the location of the object, will be still more diverged 
by the action of the lens so as to appear to proceed from f2, 
w^hich is assumed to be the location of the negative focus and is 
conjugate to fi. 

Suppose in the above diagram the lens is an 8-inch concave, 
and the distance of fi is 40 inches; then the rays enter the lens 
with a divergence of 1/40, to which is added the divergent power of 
the lens, which is 1/8, and the rays after refraction w^ill have a 
divergence 

4J "^ 8 ~ 4J " 6% 
U is negative and 6 2/3 inches in front of the lens. 

If rays diverge from 40 inches to an 8-inch concave lens, 
they are after refraction divergent as if from 6 2/3 inches. 



i 



If a patient comes to you wearing -\- 1 D. spheres for distance 
and -\r 6 D. spheres for reading, would this seem right to you? 
If not, point out where the error is likely to lie. 

We must assume that this is a case of hypermetropia with 
presbyopia, and as such the history would be more complete if 
we knew the age of the patient. However, w^e will consider the 



Practical Optometry 217 

presbyopia as complete, or in other words that the ampHtude of 
accommodation has been entirely lost. 

Under such circumstances the focal distance of the lens 
required for reading should correspond to the desired reading 
distance, as, for example, with entire suspension of accommoda- 
tion a -f 2.50 D. lens would afford a reading distance of 16 
inches, a + 3 D. lens, a reading distance of 13 inches, and so on. 

The latter has come to be regarded as a proper reading 
distance, and therefore, the strength of the lens required even 
in total presbyopia is seldom more than 3 D. or perhaps + 3.50 D. 

When we come to consider this case we see that there is an 
addition of 5 D. apparently to correct the presbyopia, which, 
in view of the above, is too much. Xow, then, it is probable 
that the + 6 D. lenses are necessary to afford clear vision at 
the customary reading distance, and therefore we would infer 
that instead of the reading glasses being too strong, the distance 
glasses are too weak. 

This error is probably due to the carelessness of the optom- 
etrist who made the examination. He, perhaps, placed the 
+ 1 D. lenses before the patient's eyes, which caused a marked 
improvement in distant vision and which the patient may have 
declared splendid. And on such declaration the glasses were 
prescribed, instead of following the rule to find the strongest 
lenses that afforded good distant vision. 

We think a further examination of such a case would show- 
that the hypermetropia was probably equal to 3 D. with an 
addition of a similar amount needed for reading. 



Pathological Optometry 

Define paresis and paralysis. Tell how you would diagnose 
each condition. 

The words paresis and paralysis are often used inter- 
changeably, but usually the former is understood to apply to 
milder cases. Paralysis is applied to that condition where there 
is a loss of power of motion or sensation, and paresis to that 
condition in which there is a slight form of paralysis. 

If there is paralysis in the internal rectus there would be a 
total loss of the power to turn the eye inwards. The eyes may 
assume a position of parallelism, but there is more likely to be 
a slight divergence, which would result in crossed diplopia, which 
is increased by trying to converge or look inwards, and diminishes 
or disappears when looking outwards or away from affected 
muscle. 

In paresis the limitation of movement would be less notice- 
able and the strabismus and diplopia would be less observable, 
perhaps only detectable when extreme convergence is attempted. 



■ What is the difference between conjunctivitis and keratitis? 

The first is an inflammation of the conjunctiva and the 
second of the cornea. 

The symptoms of conjunctivitis are congestion of the ocular 
palpebral conjunctiva, a burning and scratching as if sand was 
in the eye, a discharge— at first watery, afterwards muco- 
purulent; vision but slightly impaired, photophobia not very 
marked, slight swelling of the lids. 

The symptoms of keratitis indicate great irritability. 
Marked photophobia, excessive lachrymation and circumcorneal 
injection. Vision blurred on account of infiltration of the layers 
of the cornea. Blood vessels are seen in the cornea. Pain is 
considerable. The inflammation may extend to the iris (iritis) 
and even to the ciliary body and choroid. 

218 



Pathological Optometry 219 

Describe hemorrhage of the retina. 

Hemorrhage may occur In any layer of the retina or in any 
portion of its surface. Its color is at first bright red, later it 
becomes darker, and may finally be entirely absorbed. 

The most common seat of retinal hemorrhage is in the region 
of the disk and next in the neighborhood of the macula, the 
latter disturbing vision more and more as the macula is ap- 
proached. Hemorrhages occur mostly between the fibers of the 
inner layer and present a fiame-shape appearance. Those which 
occur in the outer layers are more apt to be round or irregular 
in shape. 

A hemorrhage may sometimes be so profuse as to break 
through the hyaloid membrane and escape into the vitreous. 

The causes of retinal hemorrhage are degeneration of the 
walls of the blood vessels as found in advanced life, heart disease, 
sudden reduction of tension, as after the operation for glaucoma, 
pernicious anaemia, gout, etc. 



What is complete dislocation of the lens called? 

In complete dislocation of the lens the optical condition is 
one of aphakia; that is, high hypermetropia and entire loss of 
accommodation . 

Name a disease of the eyes 07i account of which immigrants 
are refused admission to this country. Give reason. 

Trachoma, because of its contagiousness, due to the presence 
of a micro-organism in the discharge. It has a tendency to spread 
in certain countries, among certain races and in crowded institu- 
tions. Bad air, overcrowding, poor food and filth contribute to 
its development in connection with the contagion. 



What is pterygium? 

Pterygium is a vascular thickening of the conjunctiva, 
triangular in shape, on the nasal side of the cornea, with the 
apex of the growth towards the latter. Sight is not impaired 
unless the pterygium extends over the pupillary region of the 
cornea. The treatment, if any is required, is surgical. 



220 State Board Examinations 

What is color blindness? 

Complete color blindness would mean an entire absence of 
the color sense in which case a landscape would look like an 
engraving — a picture entirely colorless. This, however, is very 
rare. 

Partial color-blindness, as it is usually encountered, is due 
to a loss of sensation for one of the three fundamental colors, 
red, green and violet. Red color-blindness is the most common. 
This defect is found most frequently in men and is doubtless to 
some extent a matter of education. 



What is conical cornea? 

A diseased condition in which the cornea is altered in 
shape and bulges forward in the form of a cone. This is due to 
a gradual atrophic process in the central part of the cornea, as 
a result of which the intraocular tension causes it to bulge. 
The transparency of the cornea is not much impaired. The 
condition is easy of diagnosis because it is evident on mere 
inspection. 



What is color blindness and what effect does it have on visual 
acuity? 

Color blindness is an inability to distinguish between certain 
colors. As a rule, it does not impair the visual acuity, nor in- 
terfere with reading, writing or the ordinary occupations of life. 



\ 



Of what does the operation for cataract consist? 

In the removal from the eye of the crystalline lens, which 
has lost its transparency and become opaque. 



What causes a posterior staphyloma? 

This is a bulging backward of the eyeball and occurs in 
high myopia. In these cases there is usually an increased 
pupillary distance, which in connection with the close position 
at which work is done imposes an unnatural tax upon the con- 



Pathological Optometry 221 

vergence. This latter pressure from tension of the recti muscles 
tends to an elongation of the ball, which is facilitated by the 
softening of the tissues of the posterior half of the eye, due to 
the accompanying scleritis and choroiditis, which is so common 
in myopia. 

What is aphakia? 

A condition of the eye in which the crystalline lens is absent, 
either naturally or artificially, as in the operation for cataract. 



What is glaucoma, and name one prominent symptom? 

A disease which seriously threatens vision on account of 
abnormal intraocular tension, and its one prominent symptom 
is a stony hardness of the eyeball. 



What part of the eye is affected in conjunctivitis? 
The mucous membrane called the conjunctiva. 



What part of the eye is affected in iritis? 
The iris. 

What would lead you to believe that your patient was siffering 
from cataract? 

The oncome of second sight would at once lead to the sus- 
picion of cataract. In such cases there is a condition of refractive 
myopia, due to a softening and swelling and, as a result, an 
increase in power of the crystalline lens. This is apt to occur 
.in the 60's, and such a person who has been depending upon 
convex lenses fof reading now finds he is able to read without 
glasses. This is almost invariably a premonitory symptom of 
cataract. 

Also if smokiness of vision and spots before eyes were com- 
plained of cataract would be suspected. In advanced stages of 
the disease cataract is easily detected even in ordinary daylight 
by the grayish appearance of the pupil. In the earlier stages 



222 State Board Examinations 

cataract is discovered by an ophthalmoscopic examination, the 
opacities showing as dark spots on the red background. 



What is meant by the expression ''woolly ' appearance around 
the edge of the optic nerve? 

This term is used to express the appearance of the disk in 
papillitis or inflammation of the optic nerve head, in which the 
disk is swollen, red and with a blurring of its border by a grayish 
opalescent haze. This haziness may become a decided opacity, 
covering and extending beyond the border of the disk, and 
later pass into a condition of atrophy. 



A person cannot see distant objects clearly. What are the 
possible reasons? 

In a general way we say the impairment of distant vision 
may be due either to refractive error or disease. 

Any error of refraction, myopia, hypermetropia or astigma- 
tism may be the cause of it, especially myopia and high degrees 
of hypermetropia and astigmatism. 

In regard to diseases there may be cataract or opacity of 
any of the refracting media and affections of the optic nerve and 
retina. 

What is the cause of strabismus in young children? 

During the first few months of life the movements of the 
eyes are uncertain and are not controlled by the brain. It is 
not until the end of the first year that binocular vision is estab- 
lished. In the absence of the natural desire for binocular vision- 
there is no incentive for accord between the movements of the 
two eyes, under which circumstances any slight cause may upset 
the equilibrium of the higher centers of the brain. Hence babies 
often squint from gastric or other disturbances. 

But in the great majority of young children where the 
strabismus is of the convergent concomitant form and appears 
by the fourth year, it is due to hypermetropia, which is the 
predominant error of refraction in children and which imposes 
an extra tax on the accommodation, and thus stimulates the 
convergence. 



Pathological Optometry 223 

What is mixed astigmatism and what kind of a lens is required 
to correct it? 

That condition in which one meridian is hypermetropic and 
the other meridian at right angles is myopic. It is corrected by 
a cross-cyUnder, the convex element of which corrects the 
hypermetropic astigmatism, and the concave element the 
myopic astigmatism. The axes of the cylinders are at right 
angles to each other, and this cross-cylinder may be transposed 
to a sphero-cylinder. 

The far point of a myope is at 10 cm. and his near point is 
at 5 cm. What is his myopia and what is his accommodation? 

His myopia as estimated by his far point is 10 D. and his 
amplitude of accommodation as shown by his near point is 20 D. 



What is meant hy the terms astigmatism with the rule and 
astigmatism against the rule? 

Usually the normal eye shows a slight excess of curvature in 
the vertical meridian of the cornea; hence in any case of astigma- 
tism where the vertical meridian exceeds in refractive power we 
term it "with the rule," and where the horizontal meridian is 
in excess, "against the rule." 



What kind of astigmatism cannot he satisfactorily corrected 
with glasses, and why? 

Irregular astigmatism, because in these cases there is a 
difference in refraction in different parts of the same meridian, 
and it is obvious that glasses could not be ground in such a way 
as to correct this condition. 



What kind of strabismus, if any, is usual with pronounced 
hypermetropia? 

Convergent concomitant strabismus, on account of the close 
relation existing between accommodation and convergence. In 
hypermetropia where the accommodation must be used in 
excess of normal, the convergence is stimulated into excessive 
action and manifests itself by a turning in of one or the other eye. 



224 State Board Examinations 

The far point of a certain eye is 33 cm. behind the eye, and the 
near point is 33 cm. in front of the eye. What is the character of 
the refractive error, and what is the amplitude of accommodation? 

Hypermetropic to the extent of 3 D. and an amplitude of 
accommodation of 6 D. to overcome the hypermetropia and show 
a near point of 33 cm. 

The far point of an eye is 1/3 of a meter in front of the eye. 
What is the kind of error, and what lens is required to make this eye 
emmetropic? 

Myopia, requiring a — 3 D. to make it emmetropic. 



An eye is hypermetropic 2 D. in the vertical meridian and 2.50 
D. in the horizontal. Write four different prescriptions of glasses 
to fully correct the error of refraction. 

+ 2 D. cyl. axis 180° C + 2.50 D. cyl. axis 90° 
+ 2 D. sph. C + .50 D. cyl. axis 90° 
-f 2.50 D. sph. C - .50 D. cyl. axis 180° 
Toric + 6 D. vertically + 6.50 D. horizontally with — 4 D. 
for inner surface. 



An old person has no accomodation left. To see distinctly at 
33 cm. he required + 2.25 D. sphere. What would he the prescrip- 
tion for distance and for reading music at 80 cm.? 

In order to see distinctly at ?i?) cm. without effort of accom- 
modation a + 3 D. lens would ordinarily be required; but as in 
this case it was possible with a + 2.25 D. lens, we must assume 
that the other .75 D. is supplied by the eye, which could be done 
only if the eye was myopic to that extent. 

For vision at 80 cm. without accommodation a + 1.25 D. 
lens would be called for, which in this case, where the refraction 
is myopic, .75 D. would be reduced to + .50 D. 

Therefore the prescriptions would be as follows : 

Distance - .75 D.; at 80 cm. + .50 D. 



In cases of myopia when muy this condition he expected to 
show itself? 



Pathological Optometry 225 

During school life, when the eyes are taxed to the utmost 
perhaps with insufficient light and under unfavorable conditions, 
the first indication being a tendency to hold the book close to the 
e\'es. 

What is the far point of an eye ivhich needs a — 6 D. to make it 
emmetropic? 

About six and a half inches. 



What is the difference befweefi latent strabismus and hetero- 
phoriu? 

Xo practical difference, as both terms are used to express the 
same condition. 

In the case of anisometropes one retinal image may he larger 
than the other, and, therefore, the two eyes cannot act together. How 
can we settle this point? 

By placing before each eye its proper correction and directing 
the patient's attention to a single line of letters across the room. 
A 5° prism is placed vertically before one eye and the patient will 
see two lines of letters, one under the other, one belonging to each 
eye. The patient is asked which is the plainer, the upper or the 
lower. In this way can be quickly determined which eye has the 
best vision and an effort made to improve that of the poorer eye 
so as to place them as nearly as possible on an equality. 

The general rule in anisometropia is to take care of the good 
eye and give it its proper correction, with an approximate cor- 
rection to the other eye for the time being, which can later be 
raised more and more to full correction. 



In a case showing hy the trial case test a full correction with 
— .50 spheres, what would lead you to suspect that there was a spasm 
of accommodation? 

Inasmuch as hypermetropia is the predominant error of 
refraction, and as its detection is important because it is the chief 
cause of eyestrain, it is proper to suspect this error in all cases 
until proved otherwise. Slight degrees of myopia are few, and as 
weak concave lenses are readily accepted by all young people such 
acceptance should at once raise the suspicion of spasm of accom- 



226 State Board Examinations 

modation. and then the fogging system must be patiently and 

persistently used. 

What is the cause of presbyopia and when does it begin? 

Due to the increasing firmness of the crystalline lens, on 
account of which it is no longer able to respond to the efforts of 
the ciliary muscle to cause it to assume an increased convexity. 
As a result the near point commences to recede from the tenth 
year, but presbyopia cannot be said to begin until the near point 
has receded beyond eight inches, as until that time near vision 
is but little interfered with. 



What causes the recession of the near point? 

Lessening of the amplitude of accommodation, as naturally 

occurs in presbyopia. 



What tii'o fundamental reasons are there for the oncoming oj 
presbyopia? 

Weakening of the ciliary muscle and increase in firmness of 
the crystalline lens, so that it can no longer respond to the action 
of the muscle in the effort to augment its convexit^^ 



If there is a tendency of the visual axes of the two eyes to diverge, 
what is the condition called? Suppose the deviation actually exists, 
then what is the name of the condition? 

Tendency to divergence, exophoria. Actual divergence, 
exotropia. 



When a person of thirty years of age, who is an emmet rope or 
close thereto, has to wear glasses for near vision, what condition is 

probably present? 

When a person at this age begins to need convex lenses for 
reading we at once suspect hypermetropia. But as the question 
intimates a condition of approximate emmetropia the condition 
must be one of deficient or subnormal accommodation. 



Pathological Optometry 221 

When the retinal images of an object on the two retinas fall on 
such retinal points that the two images cannot he fused into one, what 
is the result and what is the condition called? 

Double vision or diplopia ; and if there is a deviation of one 
eye, strabismus. 

Define anaphoria, cataphoria and hyperphoria. 

Anaphoria is that condition in which there is a tendency for 
the eyes to turn above the normal position, as a result of which 
the head must be bent forward. 

Cataphoria is that condition in which there is a tendency for 
one visual line to deviate below the other, and in hyperphoria 
above the other. 

What name is given to that condition where one eye is hyper- 
metropic or myopic and the other eye is more so? When one eye is 
hypermetropic and the other is myopic? 

Anisometropia is the general term in use, although the latter 
may be called antimetropia. 



The punctum remotum of a patient twenty years old is seven 
inches and vision is improved by a concave lens. What condition 
exists? 

Myopia of approximately 5.50 D. 



How could you objectively measure the deviation of a squinting 
eye? 

One of the most reliable methods is by means of the arc of a 
perimeter. The squinting eye should be in the center of the arc, 
while the fixing eye should be engaged with some distant object 
across the room. The angle of deviation is found by moving 
a lighted candle along the arc of the perimeter until the reflection 
of the flame appears to occupy that portion of the cornea which is 
directly over the center of the pupil, when the eye of the observer 
is behind the flame. This locates the optic axis of the eye and the 
amount of deviation can be measured and read oft' the scale that 
is printed on the arc of the instrument. 



228 State Board Examinations 

How would you subjectively measure the deviation of a squinting 
eye? 

This is possible only when diplopia can be made evident. 
Ordinarily in strabismus the squinting eye is amblyopic and 
vision is monocular; but by exercise of the deviating eye to 
improve its vision and by placing a colored glass over the fixing 
eye to lessen its domination, it may become possible for the 
patient to discern two lights. When once this is accomplished 
there is no difficulty in measuring the deviation by the prism 
that is required over the squinting eye with its base opposite 
to the deviation that will fuse the two lights. 



What would lead to a suspicion of paralytic strabismus? 

When there is limitation of movement of eye in one direction. 

Diplopia, which is confined to one portion of the field and 
is increased the more the e^^es are turned in this direction. 

Secondary deviation of the sound eye exceeding the primary 
deviation of the affected eye. 

A strabismus that is noticeable only in certain positions. 



What are the conditions in the following cases: Parallel light 
is focussed on the retina; in front of the retina; behind the retina; 
part focussed in front of the retina and part behind? 

Emmetropia, myopia, hypermetropia, astigmatism. 



What is the difference in meaning of anaphoria and hyper- 
phoria, and can the two conditions be present at the same time? 

Anaphoria signifies an upward tendency of both eyes to- 
gether, the visual lines remaining parallel to each other. 

Hyperphoria signifies a tendency of the visual line of one 
eye to place itself above that of the other, thus causing a de- 
parture from parallelism of the visual lines. 

It is evident that these two conditions could not exist in 
the same case at the same time. 



Pathological Optometry 229 

What must be the character of the incident rays of light in 
order that they may focus on the retina when it is situated nearer 
than the principal focus of the dioptric system? 

The rays must be convergent as in hypermetropia, where 
they are made so by the correcting convex lens. 



What must he the character of the incident rays of light in 
order that they may focus on the retina, when it is situated beyond 
the principal focus of the dioptric system of the eye? 

The rays must be divergent as in myopia, where they are 
made so by the correcting concave lens. 



What is the condition of hypermetropia ivhen the accommodation 
conceals a part of the error, and what term is applied to the part 
of the error not so concealed? 

Latent when concealed by the accommodation and manifest 
when not concealed. 

- .12 D. sph. = + .25 D. cyl. axis 140° 



What will be the refractive condition of the two principal 
meridians of an eye requiring the above? 

.12 D. hypermetropic in the 50th meridian and .12 myopic 
in the 140th meridian. 

What kind of astigmatism will cylinders correct, and what kind 
will they not correct? 

In regular astigmatism in which there are two principal 
meridians at right angles to each other, one of least and the other 
of greatest refraction, cylinders supply the needed power in the 
desired meridian to produce equalization. 

But no glass can be ground to correct irregular astigmatism, 
in which there is a difference in power in different parts of the 
same meridian. 

Name the two spherical errors of the eye and state how they are 
corrected? ^ 



230 State Board Examinations 

Hypermetropia, in which there is assumed to be a deficiency 
of refractive power, and is corrected by a convex lens to supple- 
ment the deficiency; and myopia, in which there is an excess of 
refractive power and is corrected by a concave lens to reduce 
the excess. 

What is the effect of hypermetropia on the accommodation? 

In hypermetropia uncorrected, there is a constant tax or 
strain on the accommodation to make up for the deficiency of 
refractive power, causing symptoms of asthenopia, as the 
accommodation is called upon to do more work than in the 
emmetropic eye. 

What is the effect of myopia on the accommodation. 

In myopia there is an overplus of refractive power, on which 
account the eye is adapted for near vision, and the accommoda- 
tion is called upon to do much less work than in emmetropia. 



What is meant by the term spasm or cramp of the accom- 
modation? 

A persistent contraction of the ciliary muscle, which fails 
to relax, even when there is no need for its contraction. 



What is the difference between subnormal accommodation and 
subnormal range of accommodation? 

By subnormal accommodation the writer would understand 
that the power or amplitude of accommodation was below the 
normal standard for that particular age. 

By subnormal range of accommodation we would under- 
stand that the distance between near and far points is less than 
normal. 

What is meant by an ''error of refraction' of the human eye? 

The refraction of the eye is its optical condition or its action 
on the rays of light that enter it. When refraction is normal 
parallel rays are focused on the retina with the accommodation 
at rest. Any departure from this normal refraction, when the 
rays are not so focused, is known as an error of refraction. 



Pathological Optometry 231 

What kind of ametropic eyes have normal vision or better? 

Cases of simple hypermetropia of not too high degree. 
Here the accommodation is easily able to overcome the defect, 
and in connection with the contraction of the pupil that is 
sympathetically produced, the vision is thereby made good, 
sometimes even sharper than normal. 



Why are cases of anisometropia usually hard to fit? 

In favorable cases where the difference between the two 
eyes is not great, the correcting lenses afford good sight in both 
eyes and comfortable binocular vision is the result. But in 
many cases the effect of correction is a source of annoyance and 
confusion. 

The effect of the two different lenses on the apparent size 
of objects must be considered. For instance if one eye is hyper- 
metropic and the other myopic, the convex lens for the first eye 
causes an apparent increase in the size of the object, while the 
concave lens for the second eye makes the object appear smaller. 
While each of these retinal images may be distinct, yet on account 
of the difference in their size there is difficulty in fusing them, or 
if fused it is at the expense of strain. 

If one crystalline lens has been removed for cataract, on 
account of the strength of the lens that w^ill be required (+ 10 D. 
or over), the patient is oftentimes more comfortable if the aphakic 
eye be left uncorrected, that is provided the other eye has fair 
vision. 

Another source of trouble is the prismatic eff'ect of the dif- 
ferent lenses. In converging and looking through the lenses to 
the inside of the optical center, the convex lens will show a pris- 
matic effect base out and the concave lens a prismatic effect base 
in. But this disturbance is the more noticeable when looking 
above or below; the optical centers of the lenses than sideways. 

The older the patient the more discouraging is the effort to 
correct anisometropia. 

Define three kinds of asthenopia. 

Accommodative asthenopia, due to an overtaxing of the 
accommodation as in hypermetropia. 



232 State Board Examinations 

Muscular asthenopia, due to an overtaxing of the conver- 
gence as in myopia. 

Retinal asthenopia, where there is hyper-sensitiveness of the 
retina and suffering is caused by exposure to light. 



How may astigmatism of one kind he changed into that of 
another? 

In simple hypermetropic astigmatism, spasm of accommoda- 
tion transposes the case into one of apparent simple myopic 
astigmatism with the defect in the meridian at right angles. 

Then, too, with the advance of age the addition of the neces- 
sary convex lens to correct the presbyopia, changes a case of 
simple hypermetropic astigmatism and a case of simple myopic 
astigmatism into one of mixed astigmatism. 



How do some people wrongly understand the word squint? 

The terms squint and strabismus are used to describe the 
same condition, but the popular understanding of squint is the 
act of half closing the lids in the effort to see or to protect the 
eyes from excessive glare. 

What is the difference between tonic and clonic spasm of the 
ciliary muscle? 

A tonic spasm is one that is constant or continuous, while a 
clonic spasm is intermittent. 



What is the difference between concomitant and paralytic 
strabismus? 

In concomitant strabismus, the squinting eye maintains the 
same relation with the fixing eye and follows it in all of its move- 
ments. 

In paralytic strabismus, the affected eye is unable to turn in 
the direction of the paralyzed muscle, and hence cannot follow 
the movements of the fixing eye. 



When would you consider the chances good for the disappearance 
of squint as the result of wearing glasses? 



Pathological Optometry 233 

When the squint is convergent and caused by hypermetropia, 
when the error is fully corrected, and most important when the 
glasses are put on at the very commencement of the squint and 
worn constantly. 

Name the different reasons that are held in regard to cause of 
squint. 

These may be intrinsic and extrinsic. 

The intrinsic causes are error soj refraction, as evidenced by the 
convergent strabismus of hypermetropia and the divergent stra- 
bismus of myopia; and impaired vision, as shown by the strabis- 
mus that follows amblyopia and opacities of the refracting media. 

The extrinsic causes are difference in length or strength or 
innervation of the muscles; variations in pupillary distance, 
in shape of eyeball and in divergence of orbits; and impair- 
ment of fusion faculty. 

What influence, if any, has the correction of errors of refraction 
on muscular deficiencies? 

When esophoria exists with hypermetropia, the correction of 
the latter by convex lenses has a favorable effect upon the muscu- 
lar anomaly; but if exophoria is present, the convex lenses would 
aggravate it. 

If exophoria occurs in connection with myopia, the correction 
of the latter by concave lenses has a favorable effect upon the 
outward deviation; but if exophoria is present, the concave 
lenses would make it worse. 



Can a myope have a vision of 20/20 or Jaeger 1 without glasses 
under any circumstances? What would he the visual acuity of a 
myope of 2 D. and of 6 D.? 

Inasmuch as myopia always impairs the acuteness of vision, 
it is not possible for a myope to have the normal acuity of 20 20, 
but Jaeger 1 (which refers to the small size reading type) could 
easily be read if held within the far point, the higher the amount 
of myopia the more this point is restricted. 

In high myopia even with glasses of full correction, it is not 
always possible to afford a vision of 20/20, on account of the dis- 



234 State Board Examinations 

turbance to the retina from the stretching and the diminishing 
effect of the strong concave lenses. 

The visual acuity of a myope of 2 D. would probably be 
reduced to 20/200 or one-tenth, and of 6 D. to 6/200 or one-thir- 
tieth. 



An emmetrope aged 60 has had his crystalline lens removed 
for cataract. What glass would you give him for reading and what 
glass for distant vision? Could he see clearly with either glass at a 
meter? 

For distant vision, the lens needed would be about 10 D. 
For reading, inasmuch as an aphakic eye is totally devoid of ac- 
commodation, the lens must have a focal distance that will cor- 
respond with the distance at which the person desires to see. If 
this is at ten inches, the necessary addition would be + 4 D., or 
a lens of 14 D. for use at ten inches. 

Somewhat of an artificial accommodation can be produced by 
moving the lens closer to or farther from the eye. As it is moved 
farther away the effective power of the lens is increased. 

If he had a pair for distance and a pair for reading, on account 
of his loss of accommodation, neither pair would answer at one 
meter; but according to the rule herein mentioned + 1 D. would 
have to be added to his distance glasses, which would make + 11 
D. for vision at one meter. 



A child, aged 12, has vision = 20/30, but with a concave lens 
of 1 D. it is raised to 20/20. What would you suspect and what tests 
wotdd you employ to ascertain the true nature of his defect? 

We would suspect that this was not myopia at all, but proba- 
bly false or accommodative myopia due to spasm of accommoda- 
tion, and that perhaps the real condition of the refraction was 
hypermetropic. When vision is slightly impaired as in this case, 
astigmatism is also to be suspected. 

The fogging method should be used and will probably show 
if there is any latent hypermetropia and also if astigmatism is 
present. Then a careful test with the retinoscope should be made 
and a comparison of the results obtained from the two methods 
will reveal the true nature of the defect. 



1 



Physiological Optics 



Describe in your own language the human eye, giving approx- 
imate measurements of the different parts, together with the changes 
that take place in changing from distant to near vision. 

There are three coats, the sclerotic and cornea, choroid, iris 
and ciliary processes, and the retina. There are three humors, 
the aqueous, crystalline and vitreous. 

In the standard eye the nodal point is situated 7 mm. back 
of the cornea and 15 mm. in front of the retina. Allowing for 
the thickness of the coats, the anteroposterior diameter is about 
23 mm. 

The anterior principal focus is 15 mm. and its posterior 
principal focus 20 mm., as measured from the two principal 
points situated about 2 mm. back of the cornea. 

The distance between cornea and front of crystalline lens 
is 3.6 mm. 

The distance between cornea and back of crystalline is 
7.2 mm. 

The distance between cornea and center of rotation is 
13.2 mm. 

Distance between retina and center of rotation is 9 mm. 

In changing the adaptation of the eye from a far point to 
a near point the following changes take place: 

Contraction of the ciliary muscle. 

The anterior surface of the lens becomes more convex and 
approaches the cornea. 

The posterior surface of the crystalline increases very slightly 
in convexity; while the axis of the crystalline increases, the equa- 
torial diameter diminishes. The anterior chamber becomes shal- 
lower at center and deeper. at periphery. The pupillary edge of 
iris is pushed forward. The iris usually contracts, producing a 
smaller pupil. 

What is the dioptric apparatus concerned in the static and in 
the dynamic refraction of the eye? 

235 



236 State Board Examinations 

Refraction of the eye has reference to its action on the rays 
of light that enter it, whether focusing on or off the retina. If 
the word refraction is used alone, unqualified by any adjective, 
it is understood to be the static refraction of the eye — that is, 
with the accommodation at rest. When the term dynamic refrac- 
tion is used, we understand that the refractive power of the eye 
is increased by the action of the accommodation. 

In the first condition, the dioptric apparatus concerned 
would be the cornea, the aqueous humor, the crystalline lens 
and the vitreous humor, which are known as the refracting 
media of the eye. In the second condition, the accommodation 
is brought into play by the action of the ciliary muscle, but this 
really is not an addition to the dioptric apparatus, but produces 
an increase in the convexity and refractive power of the crystal- 
line lens, which is one of the media. 



Explain why the meter angle is not the same for different 
persons. 

The meter angle is the angle formed by the meeting of the 
visual line of each eye with the median line when the eyes are 
directed to an object one meter away. This angle is increased 
as the object of fixation is brought closer to the eyes. 

It is also evident that the size of the meter angle will be 
affected by the pupillary distance or the distance between the 
visual lines. The narrower the pupillary distance, the smaller 
the angle, while the wider the pupillary distance the larger the 
angle. 

What are the essential portions of the eye for carrying out the 
function of sight? 

1. A contractile diaphragm, the iris, to regulate the 
amount of light admitted to the eye. 

2. Certain refracting media, the cornea, aqueous humor, 
crystalline lens and vitreous humor, which act collectively as a 
convex lens to focus the rays of light upon the retina and form an 
image there. 

3. A contractile body, the ciliary muscle, which governs 
the convexity of the crystalline lens and acts as the adjusting 



Physiological Optics 237 

screw of a microscope, to accommodate the eye for vision of short 
distances. 

4. A fihri or plate, the retina, to receive the images of 
external objects. 

5. Communication with the brain centers by means of the 
optic nerve, which transmits the impressions formed on the 
retina. 



What would he the effect if the retina was exposed to light 
with no refracting media in front of it? 

Impressions of light would be received, but would afford no 
idea of form or outline, producing merely the sensation of con- 
fused light, amounting simply to the perception of light from 
darkness. 



What is the action of the eye on rays of light entering it? 
Explain fully how this action is performed and mention the media 
traversed. 

Passing from air (index 1.00) into the cornea (index 1.33) 
they are very strongly converged ; then continuing in the aqueous 
humor which has about the same index (1.33) and hence may be 
considered a continuation of the cornea. Passing into the 
crystalline lens which has a higher index (1.43) and with a convex 
anterior surface, the rays are still more converged but not as 
much as on entering the cornea, because the curvature is not so 
great and because the difference between the aqueous (1.33) and 
the crystalline (1.43) is not as great as the difference betv\'een air 
(1.00) and the cornea (1.33). Continuing then through the more 
refractive crystaUine (1.43) to the less refractive vitreous (1.33), 
but through a surface which is markedly convex towards the less 
refractive medium, the rays suffer more convergence and pass on 
to meet in focus upon the retina. 



Explain fully the Hehnholtz and the Tscherning theory of 
accommodation. 

The Helmholtz theory of accommodation assumes that, on 
account of the anterior extremity of the ciliary muscle being 



238 State Board Examinations 

attached to the firm sclero-corneal junction, contraction of the 
muscle will draw forward the anterior portion of the choroid; 
as a result the ciliary processes and the suspensory ligament of 
the lens would also be drawn forward, with a consequent relaxa- 
tion of the latter. 

The crystalline lens is enclosed in a contractile capsule, and 
in youth when accommodation is most active its substance is soft 
or semi-fluid. The tendency of such a gelatinous mass is always 
to approximate the shape of a sphere or globe, according to the 
laws of physics, because a fixed volume of matter presents its 
smallest area of external surface in this form, and hence the sur- 
face-tension of the mass in connection with the contractibility of 
the capsule, is constantly trying to reduce this surface area. 
Therefore under contraction of the ciliary muscle the ligament 
relaxes and allow^s the anterior surface of the lens to advance with 
a decided increase of curvature. 

Tscherning argues that accommodation is produced not by 
relaxation but by increased tension of the suspensory ligament 
through the agency of the ciliary muscle and that the increase 
of curvature of the anterior surface of the lens is confined to the 
portion near the apex of the lens, the curvature diminishing 
rapidly toward the periphery. 



What is meant by the dioptric media of the eye? 

The word "dioptric" is derived from the Greek and means to 
''look through." These are the same as the refracting media of 
the eye, the cornea, aqueous humor, crystalline lens, and vitreous 
humor, which act on and refract the rays of light entering the eye 
so as to form an image on the retina. 



What are Ptirkinje's figures? Draw a diagram illustrating the 
formation of these figures (a) when the illumination is directed 
through the sclerotic, (b) when the illumination is directed through 
the pupillary space. 

Purkinje described an entoptic method of observation of the 
retinal vessels, w^hich was intended to prove that the perceptive 
faculty for light is located in the outer portions of the cones. 



Physiological Optics 



23 > 



A strong light is focused upon the sclerotic near the cornea 
and the transmitted light will cast shadows of the intervening 
blood-vessels on the outer layer of the retina. This perception of 
one's own blood-vessles is possible because they lie anterior to the 
principal percipient layer of the retina. The disk cannot be seen 
because it has no percipient layer, but under favorable circum- 
stances the retina may be seen almost up to the disk, including 
the macula which is surrounded by capillary loops. 




Fig. Zi 

In a darkened room a candle is held to one side of the eye 
at an angle of about 30 degrees with visual line on one side, and 
if the observer's eye be at an equal angle on the other side, there 
can be seen three distinct reflected images of the candle flame. 

1. A bright virtual image of the flame reflected from the 
cornea acting as a convex mirror. 

2. A virtual, image of the flame reflected from the anterior 
surface of the crystalline lens, also acting as a convex mirror. 
This image is larger, less distinct and not so noticeable as the first. 

3. A bright but smalll-eal image reflected from the posterior 
surface of the crystalline, acting as a concave mirror. 

The first and second images are upright and the third is 
inverted. 



240 State Board Examinations 

The eye under observation is now made to change its point 
of sight from a distant to a near object; in other words, to ac- 
commodate, when it will be seen that the second image which it 
reflected from the anterior surface of the crystalline becomes 
smaller and clearer,- and moves forward and toward the center 
of the pupil; thus proving that this anterior surface of the 
crystalline becomes more convex. . Images 1 and 3 are not changed, 
thus proving that the other surfaces are not affected by accom- 
modation. 



What is the axis of visio^iF 

This is the line which joins the object with the fovea cen- 
tralis: it is also known as the visual line. 



The eye is said to be chromatic, ajid yet this condition is not 
usually manifest. Hoiv can its existence he shoum? 

By the use of a cobalt blue glass while looking at a light. 
This lens allows both blue and red light to pass through it, the 

former being- most refracted and the latter least. 



What are the 7iodal points of the eye? 

The nodal point is located at the center of curvature of the 
eye, or at its optical center, and is the point through which rays 
pass without deviation. It may be said to coincide with the 
apex of the posterior surface of the crystalline lens. 

Every lens, strictly speaking, has two nodal points, but in 
thin lenses the deviation of the secondary rays is so slight that 
for all practical purposes only one nodal point is recognized. 

The first nodal point lies 7.09 mm. behind the anterior sur- 
face of the cornea, and the second nodal point 7.46 mm. behind. 
The interval betsveen the two nodal points is so small that they 
can be regarded as one. and its distance is given as 7.2 mm. behind 
the cornea. The angle which an object subtends at the nodal 
point governs the size of the retinal image. 



Hoii' is accommodation accomplished^ 



Physiological Optics 241 

By means of the ciliary muscle acting upon the crystalline 
lens in such a way as to increase its convexity. There are two 
theories of accommodation as promulgated by Helmholtz and 
Tscherning. According to the first the lens tends to assume a 
spherical shape when allowed, which is accomplished by the action 
of the ciliary muscle on the fibers of the zonula, so as to permit 
the lens to assume a more spherical shape. According to the 
second the periphery of the lens is flattened, with the result of 
causing a bulging at the center. 



What is the optic axis of the eye? 

An imaginary line that passes through the apex of the cornea 
and the center of the crystalline lens to a point on the retina at 
the center of the fundus. 

Suppose for any reason an emmetropic eye should become 
myopic, what would he the change in the size of the apparent size 
of objects? 

The size of the retinal image is governed by the distance 
between the nodal point and the retina, that is, the distance the 
rays diverging from the nodal point must travel before reaching 
the retina. 

Inasmuch as this distance increases in the elongated eye 
of axial myopia, it follows that there will be an increase in the 
apparent size of objects. 

What is the relation of the dioptric system of the eye to the 
retina in each of the following cases: emmetropia, myopia, hyper- 
opia, astigmia? 

In emmetropia the dioptric" system is of such power as to 
focus parallel rays upon the retina with the accommodation 
quiescent. In myopia the dioptric system shows an excess of 
power, causing parallel rays to focus in front of the retina. In 
hypermetropia there is a deficiency of refractive powxr, allow- 
ing the focus of parallel rays to go behind the retina. 

In simple hypermetropic astigmatism the dioptric power of 
one meridian is just right, and of the other meridian deficient. 



242 State Board Examinations 

In simple myopic astigmatism, one meridian just right and 
the other meridian in excess. 

In compound hypermetropic astigmatism, both meridians 
deficient but one more so than the other. 

In compound myopic astigmatism, both meridians in excess 
but one more so than the other. 

In mixed astigmatism one meridian deficient and the other 
in excess. 

What position does an image on the retina have relative to the 
object? 

The reproduction on the retina of the object seen is in the 
form of an inverted image, and this image and the object are 
conjugate to each other. 

What is the dioptric value of 1 mm. variation in the distance 
of the image formed in an ey.e with the ciliary at rest? 

About 3 D. to be added if the optic axis is lengthened, and 
subtracted if the axis is diminished. 



Estimating roughly, what ivould he the increase in the power 
of an emmetropic eye if the radius of the cornea were 1 mm. more 
or less than is actually the case? 

In curvature ametropia each millimeter of lengthening or 
shortening of the radius of curvature is equivalent to about 6 D. 



In a certain eye the power of the anterior surface of the cornea 
is 42 D., while the total dioptric power of the eye is 52 D. What 
•will he the dioptric power of the eye if this head is immersed in 
water, the refractive index of the cornea being considered the same 
as water? 

Refraction occurs when light passes from one medium to 
another of different density, as long as it remains in the same 
medium there can be no refraction. In this case where the eye 
is immersed in water and where the cornea and the water have 
the same refractive index, light, in passing from the water into 
the cornea is practically traveling in the same medium, and 



I 



Physiological Optics 243 

hence there will be no refraction. The refractive power of the 
cornea being thus destroyed, there is a loss of 42 D. and the 
dioptric power of the eye is thereby reduced to 10 D. This 
explains why a swimmer who opens his eyes under the water is 
so intensely hypermetropic that he is unable to see objects 
distinctly. 

What is chromatic aberration, and hoiv is it detected in the 
human eye? 

Inasmuch as the degree of refraction by a prism \-aries with 
the wave length, therefore refraction by spherical surfaces causes 
red to deviate the least and violet the most, and as a result the 
various colors do not focus at the same point. This is known as 
chromatic aberration. 

When the eye accommodates for one set of rays, it is out of 
focus for another, which tends to cause a fringe of colors around 
the image. This defect in the normal eye is so slight that the 
brain fails to take cognizance of it and it is not noticeable in 
ordinary vision. 

That it does exist in the eye. however, can be proven by 
the use of a cobalt lens, which is placed before the eye while the 
patient is asked to look at a luminous point. This substance 
has the property of intercepting all but the red and blue rays. 
The patient will see an image with a red center clearly defined, 
and a periphery of blue, ill defined, for the reason that the eye 
more readily accommodates itself for the red rays, making their 
focal point distinct. The blue rays being more highly refracted 
come to a sooner focus, cross and diverge, giving rise to diffusion 
circles. 

If a concave lens be added to the cobalt, the blue rays will 
be focused on the retina while the red will now fall in diffusion 
circles, causing an image with a blue and distinct center and a 
red and diffused periphery. 



Name the different ametropic conditions of the eye. Explain 
each condition, giving its correction by lenses. 

Emmetropia is an eye in measure, when with a passive 
accommodation parallel rays are accurately focused upon the 



244 State Board Examinations 

retina; ametropia is a departure from this condition, showing 
itself in axial ametropia (hypermetropia and myopia) and curva- 
ture ametropia (astigmatism). Hypermetropia and myopia may 
also be due to a diminution or increase of curvature of the 
cornea or crystalline. Presbyopia is not included, because it is 
a physiological change that comes on with the advance of age. 

Hypermetropia: On account of shortness of the antero- 
posterior axis of the eyeball, or deficiency of refractive power in 
cornea and lens, parallel rays strike the retina before uniting in a 
focus, making the imaginary focus behind the retina. 

In order that light may focus upon the retina of a hyperme- 
tropic eye the rays must be convergent instead of parallel. 
There are two ways in which this can be done. 

First. By the use of the accommodation, w^hich the hyper- 
metrope does unconsciously, but as this is an unnatural use of 
the accommodation, headache and asthenopia are likely to result. 

Second. Artificially by the use of a convex lens, which 
converges the rays before entering the eye. 

Myopia: On account of the length of the anteroposterior 
axis of the eyeball, or excess of refractive power in the crystalline 
lens and cornea, parallel rays are brought to a focus too soon 
and meet in front of the retina. 

In order that the focus may be thrown farther back to the 
retina, the rays must be divergent instead of parallel. 

The eye possesses no power of its own to accomplish this, 
but it can be done artificially by means of a concave lens which 
diverges the rays before entering the eye. 

Inasmuch as the rays from near objects are divergent, they 
can be focussed on the retina of the myopic eye, hence near vision 
is clear while distant vision is blurred. 

Astigmatism. In this error there is a difference in the curva- 
ture of the cornea in its several meridians, those of greatest and 
least refractive power being at right angles to each other and 
known as the principal meridians. 

The rays passing through the vertical meridian which is 
normal, are focussed upon the retina; while the rays passing 
through the horizontal meridian which shows an excess in curva- 
ture, are brought to a focus in front of the retina. This represents 
simple myopic astigmatism, and is corrected by a concave cylin- 



Physiological Optics 245 

der, which diverges the rays passing through the horizontal 
meridian and throws them back to the retina. 

Astigmatism may be simple, where one meridian is emme- 
tropic and the other hypermetropic or myopic; compound where 
both meridians are defective, but one more so than the other; 
mixed, where one meridian is myopic and the other hypermetropic. 



What determines the visual acuity of an eye? What constitutes 
normal vision? Is normal vision always present when refractive 
conditions are emmetropic? 

The acuteness of vision is determined by the smallest dis- 
tance between two points at which they can be separately dis- 
tinguished. It has been found that the minimum visual angle for 
seeing points should be about one minute. 

Snellen's test types, which are in constant use for determin- 
ing the visual acuity, are constructed of such sizes that the whole 
letter will subtend an angle of five minutes, and each limb of the 
letter (upon w^hich its visibility depends) an angle of one minute. 

The No. 20 line when placed at a distance of twenty feet will 
subtend these angles, and hence when a person can name these 
letters at this distance, we say he possesses normal visual acuity. 

The eye may be emmetropic and a well defined image formed 
upon the retina at the macula, but if this impression is not con- 
veyed to the brain by reason of any failure of the function of the 
optic nerve, or of the retina or of the centers in the brain, there 
could be no resulting vision. 



What is the cause of presbyopia? State the functional dis- 
turbances in presbyopia. 

The cause of presbyopia is a gradual lessening of the power 
of accommodation. This is a physiological condition, and is 
dependent upon a diminished elasticity of the crystalline lens 
and a loss of contractibility of the ciliary muscle. When the 
amplitude of accommodation falls below 4.50 D. or 5 D. there is 
not sufficient reserve to allow of comfortable vision at reading 
distance, and artificial assistance is called for. 

The loss of accommodation in presbyopia destroys the rela- 



246 State Board Examinations 

tion that had in earHer years existed between the functions of 
accommodation and convergence. 



Is the refraction of the eye affected hy spasm of the ciliary 
muscle? What is (a) tonic spasm of accommodation, (b) clonic 
spasm of accommodation? 

Spasm of the ciliary muscle causes parallel rays of light to 
come to an earlier focus, thus simulating myopia. Even in hyper- 
metropia there may be an apparent myopia, caused b}^ the action 
of the muscle going beyond the limit and producing an over- 
correction. 

It is to be understood that the refraction of the eye is not 
actually altered; the change is only apparent, not real. 

A tonic spasm is one that is persistent and continuous, while 
a clonic spasm is intermittent. 



What causes diplopia? Define and explain three kinds of 
diplopia. 

When the image is focused upon portions of the two retinae 
that do not correspond, the brain is unable to fuse them, and 
double vision or diplopia results. Diplopia may be produced 
artificially by placing a prism before one eye, which deflects the 
rays from the macula of that eye, for which purpose the prism 
must not be so weak as to be easily overcome by a contraction of 
the muscles. 

Homonymous diplopia is that form in which the right image 
is seen by the right eye and the left image by the left eye. 

It is caused by an inward deviation of the eye, as in esophoria. 
In such a case the ray of light impinges upon the retina at the 
inner side of the macula, and is referred in the opposite direction, 
or outwardly according to the law of projection. 

In heteronymous diplopia, the diplopia is crossed, the right 
image belonging to the left eye and the left image to the right 
eye. This occurs in those cases of muscular imbalance where there 
is an outward deviation of the eye ; under such circumstances the 
ray of light strikes the retina at the outer side of the macula, and 
is referred in the opposite direction, that is inwardly according 
to the law of projection and the diplopia thus produced is crossed. 



Physiological Optics 247 

Diplopia may also be vertical, due to hyperphoria or hyper- 
tropia. 

In right hyperphoria, the visual line of the right eye tends 
upwards in which case the ray of light entering this eye strikes 
the retina at a point above the macula and is referred in the 
opposite direction or downward. Hence the lower image would 
belong to the right eye and the upper image to the left. 

In left hyperphoria the visual line of the left eye tends above 
that of the right, under which circumstances the ray of light 
impinges upon the retina at a point above the macula (in the 
left eye) and is referred in the opposite direction or downward. 
In this form of vertical diplopia the lower image would belong 
to the left eye and the upper to the right. 



Which are the meridians of greatest and least corneal curvature 
when the concentric circles of Placido's disk appear reflected as 
horizontal ellipses? 

The greater the curvature the closer the lines of the circle 
would be brought together ; therefore in this case where the upper 
and lower borders of the circle are approximated, we would say 
the vertical was the meridian of greatest curvature, and the hori- 
zontal the meridian of least curvature. 



What is dynamic refraction? 

The refraction of the eye is its action while in a static state 
on the light that enters it and dynamic refraction is when the 
action of the ciliary muscle is added to that of the refraction 
media. 



The far point with a plus 3.50 lens in place is at 50 cm.; and 
the same eye with a plus 1.5 lens in front of it has the near point at 
20 cm. What is the refractive condition and what is the amplitude 
of accommodation? 

If the far point with a + 3.50 D. lens is located at 50 cm., 
there would be presumably an artificial myopia of 2 D., which 
would mean an overcorrection of that amount. Therefore, + 1.50 



248 ' State Board Examinations 

D. would be the lens to make the eye emmetropic showing a 
hypermetropia of that amount. 

A near point of 20 cm. would indicate an amplitude of accom- 
modation of 5 D. but if this was obtained with the assistance of 
the correcting lens of + 1.50 then the amplitude of accommodation 
of the unaided eye would be 3.50 D. 



In presbyopia is it not a fact that the radiating fibers of the 
ciliary muscles have lost a part of their elasticity; they are not so 
elastic as when young? Would this not be a good definition of pres- 
byopia? 

The approach of presbyopia is caused by the lessening of 
accommodative power on account of the inability of the crys- 
talline lens to continue to assume the increased convexity which 
was so easily possible in earHer years. So far all are agreed; 
now then as to the cause. 

While it is doubtless true that the radiating fibers of the 
ciliary muscle lose part of their elasticity this does not enter into 
the question, as it is the circular or sphincter fibers that are 
actively concerned in the function of accommodation. 

Ordinarily we say that the failure of accommodation that 
underlies presbyopia is due to a loss of contractility of the ciliary 
muscle and of elasticity of the crystalline lens; but the probabili- 
ties are that the latter is the most important factor. Doubtless 
the muscle weakens somewhat, but it is the increasing sclerosis 
of the crystalline which can no longer respond to the action of 
the ciliary muscle, that is mainly responsible for the condition 
known as presbyopia. 

What is the size of the retinal image of an object 9 cm. high and 
1.5 meters distant from the eye of a hypermetrope of 3 D.? 

The size of the retinal image varies in different eyes on ac- 
count of the varying distance of the nodal point from the retina, 
that is, on the distance which the axial rays have diverged from 
each other after lea\ing the nodal point and when they reach the 
retina. The farther these rays have to travel before reaching 
the retina the more they are separated and, therefore, the greater 
the retinal area occupied by the image. This explains why in the 



Physiological Optics 249 

elongated eye of myopia objects seem larger and in the flattened 
eye of hypermetropia things are seen smaller. 

In addition to its distance from the nodal point the size of 
the retinal image depends upon the size of the object and its 
distance from the eye. 

The size of the retinal image bears the same relation to the 
size of the object itself as the distance between the nodal point 
and the retina bears to the distance bet^veen the nodal point and 
the object. 

In an emmetropic eye the distance of the retina from the 
nodal point is 15 mm., but in a hypermetropia of 3 D. that dis- 
tance is reduced to 14 mm. 

The problem may be expressed as follows. The size of the 

image is to size of object as image distance is to object distance. 

Let X represent the size of the retinal image and substituting 

figures we have 

X : 90 mm. : : 14 mm. : 1500 mm. 

,, 90 X 14 „ ^ , 

thenx = ^-QQ = 0.84 mm. 

Multiply size of object (9 cm.) by image distance (14 mm.) 
and divide by object distance (1.5 meters) and the result is 0.84 
mm. as the size of the retinal image. 



What muscles are hroiight into play in convergence to a point 
80 inches a?id 4 inches away.^ 

In convergence at 80 inches the internal recti muscles are 
brought into action; at 4 inches they are supplemented by the 
superior and inferior recti. 

Also in simple myopic astigmatism complicated with presby- 
opia, while the concave cylinder will serve for distance, a convex 
cylinder at right angles will be needed for reading. 



What change must be made ivith incident parallel rays of light 
so as to have them focus on the retina when it is situated beyond the 
principal focus of the dioptric system? 

If the retina is situated beyond the principal focus of the 
dioptric system, or in other words if parallel rays come to a 



250 State Board Examinations 

focus too soon (as is the case in myopia) they must be made 
divergent before entering the eye so as to throw the focus further 
back upon the retina which can be accomplished by means of a 
concave lens of the proper strength. 



Explain the reason for the improvement in vision obtained by 
the pinhole disk. 

In the several forms of ametropia parallel rays of light can- 
not focus upon the retina and as a result the acuteness of vision is 
impaired. In hypermetropia the rays reach the retina before 
coming to a focus, w^hereas in myopia they meet in front of the 
retina and cross and diverge; in both cases the retina receives 
circles of diffusion. The pinhole disk cuts off the peripheral rays 
and allows only the more central ones to pass, and in this way 
reduces the amount of diffusion and affords the more perfect 
image of the central ray which passes unrefracted. Of course, the 
illumination is reduced by cutting oft* so much light, but this is 
more than compensated for by the improvement in the sharpness 
of the retinal image. 



Explain the method applied to ascertain the powers of abduction 
and addtiction, supraduction and infraduction. 

The duction tests consist in applying prisms as strong as can 
be borne with apices over the muscle it is desired to measure. 

The strongest prisms bases in which can be overcome and 
with which single vision of a light can be maintained, will repre- 
sent the power of abduction. 

The strongest prisms bases out which can be overcome will 
represent the power of adduction. 

The strongest prisms bases doivn for supraduction, and bases 
up for infraduction. 

The eye turns tow^ard the apex of the prism, and when the 
limit of its duction power is exceeded the light becomes double. 



Why is it necessary to change occasionally the focus of glasses 
for close range after the patient has passed the age of forty? 



Physiological Optics 251 

Such persons wear glasses for the correction of their pres- 
byopia, which is an error of accommodation, and depends upon 
the lessened power of this function due to the changes which 
age brings on. As these changes are progressive the amplitude 
of accommodation gradually diminishes until it is finally lost 
altogether. This necessitates a change of glasses, increasing their 
strength about .50 D. every two and one-half 3^ears. 



What are the effects of the extrinsic muscles on the relation of 
the visual lines of the two eyes? 

The effect of the extraocular muscles is such as to keep the 
optic axes of the two eyes in a proper relation to each other in 
all the ordinary movements of the eyes that binocular vision shall 
always be present no matter in which direction the eyes may find 
it necessary to turn. 

What is the posterior principal focus of the eye? 

It is the focus for parallel rays that enter the eye, and in the 
emmetropic eye is located at the retina ; when such is the case the 
eye is adapted to receive a clear image of a distant object. 

In hypermetropia with the accommodation at rest, the prin- 
cipal focus of the eye is located behind the retina, as a result of 
which the retinal image will be blurred. In myopia the prin- 
cipal focus of the eye lies in front of the retina, and again the 
image on the retina will be blurred. 



What will he the height of the image of a 6-foot man on the 
retina of an emmetropic eye if the man is 100 feet away? 

The angle formed at the nodal point governs the size of the 
image on the retina, or we may express it in this way, that it 
depends upon the distance between the nodal point and the retina. 
The rays meet at the nodal point and then they begin to diverge. 
The farther away the retina, the more these rays diverge, and 
therefore the greater the area on the retina occupied by the 
image. This explains why in myopia, where the retina is farther 
from the nodal point the image is larger, and in hypermetropia 
where the retina is nearer to the nodal point, the image is smaller. 



252 State Board Examinations 

In calculating the size of the retinal image it is to be remem- 
bered that its size bears the same relation to the size of the 
object as the distance of the retina from the nodal point (which 
means the distance of the image from the nodal point) bears to 
the distance of the object from the nodal point, which may be 
expressed in the following proportion: 
Size image : Size object :: Distance image : Distance object. 

Allowing 15 mm. as the estimated distance of the nodal 
point from the retina and substituting figures in this case, we 
have : 

6 feet _ 100 feet 

^ ' or 1800 mm. •* "^"^" ' or 30,000 mm. 
or X = 2 7 /so of a millimeter, which is the height of the retinal image. 



What glasses must he prescribed for those who no longer possess 
the function of accommodation? 

This question no doubt has reference to old age, in which, 
according to natural laws, the power of accommodation has been 
entirely lost. 

In these cases there is usually more or less of acquired hyper- 
metropia, due to the lessening of refractive powder as a result of 
the senile changes. This will call for correction and in addition 
another pair of glasses must be given for reading, about 2.50 D. 
or 3 D. stronger than the distance pair; or preferably a pair of 
bifocals. 

In eyes which are strictly emmetropic and in which the 
function of accommodation has been lost, the glass that is given 
must correspond with the desired reading distance; if at 13 
inches, 3 D.; if at 10 inches, 4 D., and so on. 



When is hypermetropia said to be absolute? 

When it cannot be neutralized by the accommodation, 
either because the defect is of too high degree or because the 
accommodative powder has been lost, as in old age. Under such 
conditions vision is impaired at all distances, but especially near 
vision. 



Physiological Optics 253 

When is hypermetropia said to he facultative? 

When it can be neutralized by the accommodation. His 
vision is good either with or without a convex lens. This is the 
condition usually found in early life. 



What is the difference between regular and irregular astigma- 
tism? 

Regular astigmatism is due to a toric shape of cornea or 
cr\'stalline, the curvature being regular from the minimum to 
the maximum meridians, and the defect can be readily corrected 
by cylindrical lenses. 

Irregular astigmatism is caused by unevenness of the surface 
of the cornea, there being a difference in refraction in different 
parts of the same meridian. As a result of this distortion of the 
cornea, an imperfect retinal image is formed and cylindrical 
lenses are of little value. 



What is the difference in meaning between esophoria a7id 
esotropia? 

Esophoria signifies a tendency to inward deviation, esotropia 
an actual turning of the eye inwards. 



Which will produce the poorest visual acuity, an error of 2 D. 
spherical or 2 D. astigmatically? 

If the error is myopic, it is probable that the spherical error 
will impair vision the most. But if the error is hypermetropic 
of not too high degree it is probable the accommodation can 
more completely overcome the spherical error, and hence in 
this case it is likely that the astigmatic error will produce the 
poorest vision. . 

In measuring the amplitude of accommodation with a reading 
chart why is the test not exact? 

Because for short distances the type is too large and because 
a person becomes familiar with certain words by their appearance 
as to the number of letters composing them and the shape of 



254 State Board Examinations 

such letters, so that he is able to make a very good guess at 
words even when he does not see the letters plainly. 



What is hyperphoria and what is the position of the correcting 
prisms? 

Hyperphoria is the term used to indicate a tendency of the 
visual line of one eye to place itself above that of the other. 
This upward tendency may affect either eye, and hence we have 
right hyperphoria or left hyperphoria. 

The position of the correcting prism is base down over the 
hyperphoric eye, or base up over the other eye, or divided be- 
tween the two eyes in these positions. 



When the refraction of the two eyes differs what term is applied 
to the condition? 

Anisometropia is the usual term, although the word anti- 
metropia is also used. It is seldom we find the tw^o eyes exactly 
alike, but the term is -applied only when the difference is great 
enough to receive consideration. 

Anisometropia is usually a congenital condition, but it may 
also be due to the natural progress of a myopia, to operation or 
injury, and to changes in corneal curvature. 



If the punctum proximiim in emmetropia is at 33 cm. when 
the eyes are being assisted by + 1.50 D. lenses, what is the amplitude 
of accommodation and what lenses should be prescribed for reading 
at 33 cm., so as to keep one-third of the accommodation in reserve? 

A near point of ZZ cm. represents an amplitude of accom- 
modation of 3 D. and as this is accomplished by the assistance 
of a + 1.50 D. lens the natural amplitude of accommodation 
would be 1.50 D. 

In order to read at ?>?> cm. 3 D. of power is required, and in 
order that one-third of this person's accommodation should be 
kept in reserve, he can use only 1 D. of his own accommodation 
and the balance must be supplied by a + 2 D. lens. 



Physiological Optics 255 

When the eye turns up and in, by what muscles is it controlled? 

Superior rectus and internal rectus, the upward movement 
being probably assisted by the inferior oblique. 



In the static eye in the case of an emmetrope, what will he the 
effect on the retinal image of wearing a -\- 1 D. cylinder, axis 180°? 

To distort the image, elongating it in the vertical meridian 
or at right angles to axis of cylinder. 



What is the condition of the muscles of the eyes in each of the 
following cases: Orthophoria, esophoria, exophoria, hyperphoria? 

Orthophoria, muscles properly balanced; esophoria, ten- 
dency to overconvergence ; exophoria, tendency to over-diver- 
gence; hyperphoria, a tendency of one visual line above the 
other. 

In esophoria the correcting prism is placed base out, in 
exophoria base in, and hyperphoria base down over the hyper- 
phoric eye. 

What is spasm of the ciliary muscle? 

An involuntary cramp or contraction of the ciliary muscle. 
This is the term applied to an overaction of the accommodation, 
occurring especially in young people in the effort to overcome 
a condition of hypermetropia or hypermetropic astigmatism and 
makes the eye apparently and subjectively myopic, thus leading 
the inexperienced optometrist sometimes into the error of pre- 
scribing concave glasses. 

It is more apt to occur in patients whose nervous system is 
broken down and oftentimes in those with a relatively weak 
accommodation . 

The usual symptoms are photophobia, lachrymation, pain, 
contracted pupils and congestion of the eye, together with the 
appearance of myopia. 



How is spasm cured? 

The first step in the treatment is the removal of the cause if 
it can be ascertained, as the correction of any existing hyperme- 



256 State Board Examinations 

tropia or astigmatism. Or convex lenses for fogging to induce 
relaxation of the accommodation, a weaker pair for distance and 
stronger for reading. Rest of the eyes and temporary abstinence 
from reading and sewing should be advised. 



What is diplopia and what is the explanation? 

Diplopia is double vision, two images being seen instead of 
one, and is due to the fact that the images are not formed on 
corresponding parts of the two retinae. 



What is the character of the difference in direction of the eyes 
if diplopia is homonymous? 

Esophoria, or inward deviation. 



What is monocular diplopia and what is its explanation? 

Multiple vision in one eye, usually due to a slight difference 
in the index of refraction of the several main segments of the 
crystalline lens, each of which gives rise to a separate retinal 
image. 

In emmetropia these images are so close that they are fused 
into one, but in ametropia, where the retina is not at the position 
of the average focus, vision is multiple. 



What is heterophoria and how does it differ from diplopia? 

Heterophoria is the term used to include all those conditions 
in which there is a tendency to depart from the normal muscular 
balance, which nature is able to compensate for and keep latent, 
but which is demonstrable by the usual tests. 

Diplopia means double vision, and is produced by and is a 
symptom of heterotropia, a term used to include all those con- 
ditions in which nature is unequal to the task of maintaining its 
desire for single vision. 

What is anisometropia? 

This term is applied to that condition of the eyes where the 
refraction is unequal. In a restricted sense of the word we 



Physiological Optics 257 

seldom find two eyes exactly alike, but the term is used only 
when the difference is great enough to be taken into account. 
The usual condition is where the character of the refraction is 
the same in both eyes, varying only in degree. Where the 
character of the refraction varies in the two eyes, that is one eye 
hypermetropic and the other myopic, the term antimetropia is 
sometimes used. 

What conditions produce diplopia? 

Diplopia occurs when the visual axes are directed so that the 
image of the object does not fall upon identical parts of both 
retinae. Any disturbance of the extra-ocular muscle balance, as 
in heterotropia, or even in heterophoria may cause this condition, 
and sometimes it is due to faultv innervation. 



Which surface of the cornea has the greater refractive power, 
and why? 

The refracting power of the cornea lies chiefly in its anterior 
surface. The effect of the posterior surface is neutralized by being 
in contact with the aqueous humor which has the same index of 
refraction, and, therefore, the aqueous may be considered as a 
continuation of the cornea. 



How can the degree of convergence he determined for any given 
working distance, and for any given pupillary distance? 

The rule is to divide the pupillary distance by the unit of 
displacing power at the given working distance. 

If we take a working distance of 13 inches (which is one- third 
of a meter) the displacement would be 3.33 mm., based on the 
unit or standard of displacement of 10 mm. for each meter of 
distance. We will assume the pupillary distance is 2>^ inches, 
which is equivalent to 64 mm. Then the number of degrees of 
convergence in a patient with the pupillary and working distances 
mentioned above is found by dividing 3.33 into 64, and the result 
is 19.21 degrees of convergence. 



What is the dioptric power of the anterior surface of the cornea, 
and what is its power as a convex mirror? 



258 State Board Examinations 

In order to find the refractive power of the cornea we divide 

its index of refraction, less unity, into its radius of curvature, as 

follows : 

8 mm. 



(1.33—1) 



= 24 mm. 



24 mm. is its focal length and we find its refractive or dioptric 
power by dividing this into 1000 mm., and the result is 41 D. 

Assuming, as we have done in the above example, that the 
radius of curvature is 8 mm. and as the focus of a mirror is one- 
half its radius, its focal length as a mirror is 4 mm., and its power 
is found by dividing this into 1000 mm., the result being 250 D. 



What is difference between binocular vision and fusion? 

They mostly convey the same meaning; but a person with 
binocular vision whose fusion sense is but feebly developed will 
under unfavorable circumstances abandon the effort and drop into 
monocular vision. While another person whose fusion sense is 
well developed will have such an intense tendency to binocular 
vision that nothing will cause him to abandon it. 



What is the name given to that part of the retina in which visual 
acuity is the sharpest? 

Macula lutea, or yellow spot in general and fovea centralis 
in particular. 

What would be the change in the refraction of the cornea if its 
radius of curvature should be changed about 1 mm.? 

It depends upon whether the change of radius was in the 
direction of an increase or a decrease. If the radius of curvature 
was increased 1 mrn. there would be a decrease in refractive power 
of approximately 5 D. Whereas, if the radius of curvature of the 
cornea was lessened 1 mm. there would be an increase of refractive 
power of nearly 7 D. 

What would be the change in the dioptric power of the crystalline 
lens if the radius of curvature should be changed 1 mm.? 



Physiological Optics 259 

Here the two surfaces of the crystalHne must be taken into 
account In considering its dioptric power. 

If the anterior surface showed an increase in radius of 1 mm. 
there w^ould be a loss in power of 1 D. If this surface showed a 
decrease in radius of 1 mm. there would be an increase in refrac- 
tive power of 1 D. 

If the posterior surface showed an increase in radius of 1 
mm. there would be a loss of power of 2 D. If this surface 
showed a decrease of 1 mm. there would be an increase in refrac- 
tive power of 3 D. 

If the radius of the two surfaces was changed 1 mm. the 
resultant change in dioptric power can be figured from above. 



The distance between the centers of the two eyes when they are 
looking into distance is 62 mm. If the glasses used are placed about 
12 mm. in front of the cornea and the object of regard is at a distance 
of 20 inches, what must be the pupillary distance of the glasses? 

An optical writer figures it as follows : The centers of rotation 
are 13 mm. back of cornea, therefore, the distance from these 
centers to the lenses would be 25 mm., or 1 inch. Then the lenses 
would be 19 inches from the object, and the p. d. of the glasses 
would be 19/20 of 62 mm. or 59 mm. 



At what ages does the accommodation of the eye being to decrease? 

At ten years, continuing gradually to decrease until it is 
entirely lost in old age. 



What is the difference in the function of the blind spot and of 
the yellow spot? . 

The blind spot is at the entrance of the optic nerve, and as it 
is insensible to light it cannot be said to have any function. 

The function of the retina is to receive the images of external 
objects and transmit the impressions to the brain. The yellow 
spot is that part of the retina that is most highly developed for 
this purpose. 



260 State Board Examinations 

Why is the pupillary distance less for near glasses than for 
distant ones? 

Because the eyes turn in in response to the increased con- 
vergence that is necessary to maintain binocular vision at near 
points, and in so doing the pupillary distance is necessarily 
lessened. 



What is meant hy the term function of convergence? 

When the word convergence is used some students at once 
think of the bringing to a focus of parallel rays of light, as by 
action of a convex lens; but in this case the word refers to the 
function that has control of the direction of the visual axes of the 
two eyes, so that they shall both meet at the point of fixation, 
w^hich is at some finite distance. 



What is the difference in the direction of the line of sight and of 
the optic axis of the eye? 

Inasmuch as the fovea centralis does not lie exactly upon the 
optic axis it follows that the optic axis and the visual axis, which 
joins the fovea with the point looked at, do not coincide. The angle 
formed at the nodal point where the optic and visual axes cross 
has been termed the angle alpha, or angle gamma. Its size is 
variable, in emmetropia usually not more than five degrees, 
increasing in hypermetropia and decreasing in myopia. 



What is the fundamental difference between the two leading 
theories of accommodation? 

According to the Helmholtz theory the contraction of the 
ciliary muscle causes a relaxation of the capsule of the crystalline 
lens and allows it to assume a greater convexity through its 
natural desire to become so. According to the Tscherning theory 
the contraction of the ciliary muscle flattens the periphery of the 
crystalline and causes it to become more bulging at the center. 
Under either condition there is an increase in the dioptric power 
of the lens. 



Physiological Optics 261 

What is the total length of the human eye and the dioptric value 
when in a static condition? 

Approximately 24 mm. and 58 D. respectively. 



What is meant by the term visual acuity? 

Visual acuity has reference to the image formed on the retina 
and thence transferred to the brain. It is the sharpness of sight 
and is determined at a distance of twenty feet or more with accom- 
modation at rest, and depends upon the smallest retinal image, 
the form of which can be distinguished by the brain. It is ex- 
pressed by a fraction, the numerator of which is the distance of the 
letters and the denominator the size letters that can be named. 
Visual acuity equals 20/20 means that the letters are 20 feet aw^ay 
and the patient can read the No. 20 line. 



With eyes in the primary position, what prism is needed to 
produce binocular vision for a small object at a distance of one meter 
from the eye? 

If we assume the pupillary distance to be 60 mm. then the 
visual lines will be the same distance apart. At a distance of one 
meter a prism of one degree w^ill show a displacing power of 10 
mm.; therefore, to overcome the separation of 60 mm. at one 
meter distance a prism of 6° would be required, or a 3° prism over 
each eye. 

What is the relation between diopters of accommodation and 
meter angles of convergence? 

There is a close and constant relation between the two. At 
a distance of one meter there is one diopter of accommodation 
and one meter angle of convergence, and for all distances the 
same proportion should hold good — that is, one meter angle of 
convergence for each diopter of accommodation. 



How can it be proved that the human eye is not achromatic? 

It is not possible for red and violet rays (from the two ends 
of the spectrum) to be focused on the retina at the same time. 



262 State Board Examinations 

But these colors possess little luminosity as compared with yellow, 
on which vision chiefly depends, and hence the fact that they are 
out of focus at the retina is not of great importance, nor does 
chromatism manifest itself in ordinary vision in approximately 
normal eyes. Its effects are more noticeable in ametropes, but 
its existence in any eye can be proved by looking at a light through 
a cobalt-blue glass, which blocks the central part of the spectrum 
and allows mainly the red and blue light to pass through it, the 
two extremes of the spectrum. The patient may see a red center 
with a blue border, or a blue center with a red border, depending 
on which color focuses nearest the retina. 



What causes the amplitude of accommodation to decrease? 

Principally due to loss of elasticity or increase in firmness 
of the crystalline lens, with perhaps some loss in contractibility 
of the ciliary muscle, which conditions are the natural accom- 
paniments of age. 

What is meant by positive and negative convergence? 

Positive convergence is the turning of the eyes inward from 
parallelism, as in the act of convergence, and negative con- 
vergence is turning the eyes outward from parallelism, as in the 
act of divergence. 

In what units of measurement is the amplitude of convergence 
expressed, and describe the unit? 

In meter angles, the unit of which is the angle formed by 
the visual lines of the two eyes meeting at a point one meter away. 



What is meant by the static refraction of the eye? 

The action of the refracting media of the eye on light unin- 
fluenced by the accommodation. 



What is meant by the term dynamic refraction of the eye? 

The action of the refracting media of the eye augmented 
by the power of accommodation. 



Physiological Optics 263 

Describe two ways of getting the amplitude of the accommodation. 

The usual way is by means of the small test types and noting 
the closest point at which they can be read. 

Or a card may be used in which two small holes are pricked 
by a pin, the distance between which must be less than the 
diameter of the pupil. The card is held close in front of the eye 
and a small needle viewed through the pinholes. When the 
needle is brought too close it appears blurred and double. The 
closest point at which it remains clear and single represents 
the amplitude of accommodation. 



By what test may the effect of a refractive error in the eye he 
so cut down that vision will he improved, and what is. the explanation 
of its action? 

By the pinhole test, which cuts off so many rays and nullifies 
the refractive power of the eye, allowing the image to be formed 
by the action of the pinhole alone, through which a ray from each 
point of the object passes, and as the rays pass they cross and 
form an inverted image. 

What is the nature of the retinal image as to exactness of focus 
in emmetropia, in hypermetropia, in myopia and in astigmatism? 

In emmetropia the retinal image is ideally perfect. In hyper- 
metropia, if not of too high a degree, it is approximately distinct, 
being made so by the accommodation; if the accommodation is 
not used the image would be blurred. In myopia it is indis- 
tinct except when the object is placed at the far point of the 
eye. In astigmatism the retinal image is in the shape of a line 
instead of a point, but in low degrees of hypermetropic astigma- 
tism the ciliary muscle is usually able to make vision clear. 



When the eyes are first fixed on an ohject at close range and 
then on some ohject further away what ocular changes occur? 

The ciliary muscles relax so as to cause the crystalline to 
become less convex and thus adapt the eye for distant vision; 
there is also a lessening of convergence so as to allow the visual 
axes to meet at the greater distance desired. 



264 State Board Examinations 

When the far point and near point are both positive or real 
what is the space between them called? 

The range of accommodation. 



With a -\- 1 D. lens placed in front of a certain eye the near 
point is at 20 inches. The test shows that the static refraction of the 
eye is 2.50 D. hypermetropic. What is the amplitude of accommoda- 
tion? 

If the near point is located at 20 inches by the aid of a + 1 
D. lens then the unassisted eye would show a near point of 40 
inches, which is equivalent to 1 D. of accommodation. 

However, as the 2.50 D. of hypermetropia must first be 
neutralized by the accommodation before near vision is at- 
tempted, the amplitude of accommodation must be 3.50 D. 



What is the size of the object 5 meters away that forms on the 
retina of a hypermetrope of 3 D. an image 5 mm. in size? 

The size of the retinal image bears the same relation to 
the size of the object as the distance from the nodal point to the 
retina bears to the distance from the nodal point to the object. 
In the emmetropic eye the distance between the nodal point 
and the retina is estimated at 15 mm. But in a hypermetrope 
of 3 D., as in this case, this distance is reduced to 14 mm. 

Then we have the following proportion : the size of the object 
is to the size of the image as the distance of the object is to the 
distance of the image. Substituting the figures: 
X : 5 mm. :: 5,000 mm. : 14 mm. 
14 X = 25,000 mm. 
X = 1,785 mm. 
which is the size of the object. 



How does the function of the eye differ in dynamic and static 
refraction? 

Static refraction with the fixation at 20 feet or more is 
considered a passive condition, while dynamic refraction with 
the fixation at the reading or w^orking distance is an active 
condition with the ciliary muscle at work. 



Physiological Optics 265 

How many diopters of accommodation are required in an 
emmetropic eye to see clearly an object at 20 feet? 

In the practice of optometry a distance of 20 feet is selected 
for making the visual tests, on the assumption that rays proceed- 
ing from this distance are practically parallel. But this is not 
strictly correct, because rays at this distance have a divergence 
of 1/240 of an inch, and would require a + .17 D. (one-sixth of a 
diopter) to make them parallel, or the use of an equal amount 
of accommodation. 



How do we measure the amplitude of accommodation, and 
why is it expressed in diopters? 

We measure the amplitude of accommodation by finding 
the position of the near point, which is the closest possible point 
one can see with the strongest effort of accommodation, and it 
is equivalent to a convex lens (expressed in diopters) of such 
strength as will give to rays proceeding from this near point, 
a direction as if they came from distance, or in other words makes 
them parallel. 

How is the size of the image of an object that is defined by the 
minimum visual angle at 6 meters distance determined when the 
nodal point of the eye is 15 mm. in front of the retina? What is 
the size of the objects? 

The size of the retinal image of an object bears the same 
relation to the size of the object itself as the distance between 
the ncdal point and the retina bears to the distance between 
the nodal point and the object. 

As the ncdal point in the emmetropic eye is usually estimated 
to be 15 mm. from the retina, the proportion is as follows: As 
the distance of the object is to the size of the object, so is 15 mm. 
to the size of the retinal image. 

Or in other words, the size of the object is multiplied by 15, 
and the result divided by the distance of the object, in order to 
find the size of the retinal image. 



What is the difference in the focusing arrangement of the 
human eye and the photographer' s camera? 



266 State Board Examinations 

In the eye the focusing is accomplished by an increase or 
decrease in curvature of the crystalHne, while the retina remains 
in a fixed position to receive the image. In the camera, the 
curvature and position of the lens is fixed and unchanging, while 
the screen is adjustable so that it can be moved to the position 
where it will receive the image most distinctly. 



If diplopia can he produced with a three-degree prism base in, 
what is the trouble? 

When a prism is placed before the eyes base in, the external 
recti muscle are brought into action to prevent diplopia; there- 
fore, if diplopia is produced, it must be because of deficiency in 
power of these muscles. 

What is diplopia and in what way does it differ from hetero- 
phoria? 

Diplopia means double vision. It is only a symptom, but 
it is one that is self-evident and needs no test to discover it. 

Heterophoria is a generic term used to indicate some im- 
balance of the extra ocular muscles. It is, however, a latent 
condition, and its existence is detected only by the employment 
of certain tests. 

Is the refraction of the eye affected by spasm of the ciliary 
muscle and if so in what manner? 

Spasm of the ciliary muscle by adding to the dioptric power 
of the eye changes its apparent refraction. In hypermetropia 
such a spasm conceals part or all of the error, so that not sufficient 
convex will be accepted or even concave lenses may be called for 
instead. 

In emmetropia, spasm makes the eye apparently myopic; 
and in myopia causes too strong a concave lens to be chosen. 



What must be the form of a cone of light entering the static 
myopic eye so as to be focused on the retina? What must he the form 
of cone of light entering the static hypermetropic eye so as to focus 
on the retina? 



Physiological Optics 267 

As the m\'opic eye is adapted for divergent rays, the apex 
of the cone must be at the far point of such eye. And as the 
hypermetropic eye is adapted only for convergent rays, the apex 
of the cone must be at the retina. 

In the first case the cone of divergent rays is a natural 
condition as light proceeds from close objects; in the second 
case the cone of convergent rays must be produced artifically 
by a convex lens in front of the eye. 



Supposing a person otherwise emmetropic had got a high 
degree of astigmatism, what will he probably complain of in the way 
he sees objects; in other words how will he, never having heard of 
astigmatism, describe his visual defects? How could you make an 
emmetrope see things as though he were astigmatic? 

This question is worded strangely as it is not customary to 
refer to an eye as being emmetropic if it has a high degree of 
astigmatism. 

If one meridian of an eye was emmetropic and the other 
chief meridian was highly ametropic, as for instance in simple 
myopic astigmatism, such an eye w^ould see clearly in one meri- 
dian and indistinctly in the other. 

For instance, he might be able to see clearly telegraph 
poles and at the same time the telegraph wires might be indis- 
tinguishable. Or he might be able to see the hands on the clock 
and tell the time of day at certain periods of the 24 hours and 
not at others; such cases having occurred and been reported 
before astigmatism was understood, as periodical obscuration of 
vision. 

An emmetrope can be made artificially hypermetropic 
astigmatic by placing a concave cylinder in front of his eye, and 
myopic astigmatic by placing a convex cylinder in front of his 
eye. 



If a person aged twenty is hypermetropic 4 D., how much 
accommodation would be used when reading at 40 cm. while wearing 
+ 1.75 D. sphere? How much accommodative power has he to 
spare and what glasses ought he to use? 



268 State Board Examinations 

This person must first correct his hypermetropia by the use 
of 4 D. of accommodation, and then he must use another 2.50 
D. of accommodation to see at 40 cm., which equals 6.50 D., 
but since he is wearing + 1.75 D. spheres, the effort of accom- 
modation would then be 

6.50 D. - 1.75 D. = 4.75 D. 

The average amplitude of accommodation for an emmetropic 
eye at twenty years is 10 D., but since this person must use 4 D. 
of it to overcome his hypermetropia, the amount of accommoda- 
tion available would be reduced to 6 D. If he is wearing the 
+ 1.75 D. spheres mentioned, this would be increased to 7.75 D. 

As to what glasses he should wear, this is a question that 
cannot be answered offhand. Ordinarily, it is not wise to correct 
more than the manifest error in any particular case, and this is 
a variable quantity. We presume that the + 1.75 D. spheres 
he is wearing represent the manifest hypermetropia and unless 
there are some special symptoms we would not consider a change. 
If, however, there were urgent symptoms calling for relief, 
then we would increase the lenses even to the extent of slight 
fogging. ^ 

Describe the crystalline lens from an optical point of view 
omitting all reference to its minute anatomy. What would he the 
optical effect of removing the crystalline lens altogether? 

The crystalline lens is a bi-convex lens, about 8 mm. wide 
and 3.6 mm. thick. The nucleus has a higher index of refraction 
than the cortex, and hence the crystalline may be regarded as a 
biconvex lens of high power enclosed between two convexo- 
concave lenses of lower power, and as a result of this construction, 
the refractive power of the lens is greater, the alteration in 
curvature in the act of accommodation is made possible and 
spherical aberration is reduced. 

The radius of curvature of the anterior surface of the 
crystalline is about 10 mm. and of the posterior surface 6 mm. 

The index of refration of the crystalline is about 1.44, but on 
account of the media on both sides of it having an index of 1.33, 
the relative index of refraction of the crystalline lens is expressed 
by 

1 44- 

T33 = 1-08 



Physiological Optics 269 

Its refractive power is from 15 D. to 20 D. 

The removal of the crystalHne makes the eye hypermetropic 
to the extent of only 10 D., which is not as much as indicated by 
the power of the crystalline lens in the eye, where its effect is 
lessened by being in contact with media approximating its own 
index of refraction. 

What do you understand by the terms ''amplitude of accommo- 
dation" and ''range of accommodation?'' Illustrate these terms in 
the case of an emmetrope aged 20, a myope of 3 D. aged 20 and a 
hypermetrope of 3 D. of the same age. Where ivoidd the near point 
be situated in each case? 

The amplitude of accommodation in all the cases mentioned 
at 20 years of age is 10 D.. but the positive refracting power and 
the near point vary in each case. 

In emmetropia the positive refracting power is identical with 
the amplitude of accommodation and the near point is 4 inches. 

In a hypermetrope of 3 D. the positive refracting power is 
reduced to 7 D. and the near point recedes to 5^ inches. 

In a myope of 3 D. the positive refracting power is increased 
to 13 D. and the near point approaches to 3 inches. 



In the application of the laws of conjugate foci to the human 
eye, what two points are conjugate? 

One conjugate focus is on the retina where the distinct image 
must be formed. The other conjugate focus will be the object 
which emits the rays that go to form the retinal image. 

Or we may say that the far point and the retina are conjugate 
to each other. 

The standard position of the retina is at the posterior princi- 
pal focus of the refracting system of the eye, which is the focus 
for parallel rays proceeding from infinity. 

In emmetropia the retina and infinity are at conjugate foci. 

In myopia the retina lies beyond the principal focus; and in 
hypermetropia it lies in front of it. 

When the retina is situated beyond the principal focus as in 
myopia, the em.erging rays will be convergent and focus at its 
far point, which is then conjugate to the retina, these conjugate 
foci being on opposite sides of the refracting system and the image 
is negative. 



Anatomy of Eye 

Name the intrinsic and extrinsic muscles of the eye and the 
nerves supplying innervations to each. 

The intrinsic muscles of the eye are : 

{a) Ciliary muscle, supplied by the third cranial, or oculo- 
motor nerve. 

{h) The muscles of the iris, of which the sphincter is supplied 
by the third cranial nerve and the dilator by the sympathetic. 

The extrinsic muscles are: External rectus, supplied by the 
sixth cranial nerve; internal rectus, superior rectus, inferior rec- 
tus, inferior oblique, supplied by third cranial nerve; superior 
oblique, supplied by fourth cranial nerve. 



Name the hones of the orbit. 

Frontal, sphenoid, ethmoid, superior maxillary, palate, malar 
and lachrymal bones. 

Describe the iris and name its functions. Name the two muscles 
which control its movements. Name the nerve supply of each. 

The iris is the terminal portion of the choroid, and is sus- 
pended in the aqueous humor in front of the crystalline lens. Its 
free border encircles the pupil and rests on the anterior surface 
of the capsule of the lens. The diameter of the iris is about 1 1 mm. 

Its vascular stroma layer is composed of bundles of loose 
connective tissue and contains blood-vessels, nerves, lymph 
spaces, nucleated cells with pigment and the muscles of the iris. 

The two muscles are the sphincter, which is a flat band 1 mm. 
in width, lying near the pupillary margin, and the dilator, whose 
fibres are arranged meridionally, extending from the ciliary mar- 
gin of the iris to the pupil. Both muscles are involuntary. The 
former is supplied by the third nerve and the latter by the sym- 
pathetic nerve. 

270 



Anatomy of Eye 271 

The iris is a diaphragm, changing the size of the pupil so as 
to regulate the quantity of light entering the eye, the movements 
being made unconsciously. These movements are induced reflexly, 
as by light, and directly, as by the accommodation. 



Describe the crystalline lens and name and describe its functions . 

This is a transparent lenticular body, classed as biconvex, 
(the posterior surface showing the greatest convexity) , resting in 
a depression in the anterior surface of the vitreous humor and 
supported by the suspensory ligament, or zone of Zinn. Early 
in life the lens substance is soft and of the same consistence 
throughout, but gradually the central portion hardens and in 
advanced age forms the nucleus. This hardening results in loss of 
elasticity and is made manifest by impaired accommodation. 
The lens substance is enclosed within an elastic strong membrane, 
called its capsule. The lens measures about 10 mm. transversely 
and 4 mm. in thickness. 

The substance of the lens is composed of layers of elongated 
cells, called the lens fibers, which are united by a cement sub- 
stance. 

The function of the crystalline lens is to increase and de- 
crease in convexity and thus provide for focusing upon the retina 
the rays of light proceeding from objects situated at different 
distances within infinity. During the act of accommodation the 
lens becomes more convex, especially upon its anterior surface. 
This increase in convexity is brought about by the action of the 
ciliary muscle acting upon the suspensory ligament. 



Describe the anterior chamber of the eye. 

The cavity in the anterior portion of the eye occupied by the 
aqueous humor is divided by the iris, which stretches across it 
from side to side, into two portions, called the anterior and 
posterior chambers. The. anterior chamber is, therefore, that 
portion of the aqueous cavity lying in front of the iris and is 
bounded anteriorly by the cornea. 



Which is the longest diameter of the eyeball? Why? 



272 State Board Examinations 

The anterio-posterior diameter is the longest because of the 
projection of the cornea. 

In high hypermetropia, it is conceivable that this diameter 
may be shorter. 

Give the location of the lachrymal gland. Describe briefly the 
lachrymal duct. 

The lachrymal gland is a compound racemose gland of the 
size and shape of an almond and is lodged in a depression at the 
upper and outer portion of the orbit. Its concave under surface 
rests upon the globe of the eye, with the conjunctiva, superior 
and external recti muscles intervening. It is held in contact with 
the orbital periosteum by a few fibrous bands. Connected with 
the gland there are a number of ducts, from eight to twelve, which 
open by minute orifices in a row on the upper and outer part of the 
conjunctival reflection. 

What is probably meant by the latter part of the question is 
a description of the nasal duct. This is a membranous canal about 
3/4 of an inch long, extending from the termination of the lach- 
rymal sac through the osseous nasal duct to the inferior meatus of 
the nose, passing in a direction downward, backward, and out- 
ward, its diameter being narrowest at its middle. Externally it is 
composed of fibro-areolar tissue and internally of mucous mem- 
brane, continuous with that of the nose and lachrymal sac. 



Name the humors and the refracting media of the eye. 

The humors from before backward are the aqueous humor, 
crystalline lens, and vitreous humor, which together with the 
cornea, form the refracting media of the eye. 



How many muscles are there in each orbit? Give their names. 

Answered above, to which, however, may be added the 
levator palpebrae superioris, which arises from the lesser wing of 
the sphenoid bone and passes forward to be inserted into the 
upper border of the superior tarsal cartilage and the skin. 



What are the principal blood-vessels of the eye? 



A natomy of Eye 273 

The ophthalmic artery enters the orbit by the optic foramen 
and gives off the following branches: 

Lachrymal, supraorbital, superior and inferior palpebral, 
nasal, long, short and anterior ciliary, arteria centralis retinae 
and infraorbital. 

Name the nerves of the eyeball. 

Optic nerve, the special nerve of sight. The motor nerves are 
the third, fourth and sixth cranial. The ophthalmic division of the 
fifth cranial nerve supplies general sensation. The sympathetic. 
The long and short ciliary. Lachrymal, frontal and orbital. 



What are the folds of the conjunctiva called? 

The folds formed by the passage of the conjunctiva from 
the lids to the eyeballs are called the superior and inferior pal- 
pebral folds, the former being the deeper. The word fornix is also 
used. 



What are the appendages of the eye? 

The appendages of the eye are the eyebrows, or supercilia; 
the eyelids, or palpebrae, with their lashes; the mucous membrane, 
or conjunctiva, and the lachrymal apparatus, consisting of the 
lachrymal gland, with its ducts, the canaliculi, the lachrymal sac 
and the nasal duct. 

Describe the posterior chamber of the eye. 

This term may have two meanings. It is possible that the 
examiners may have had reference to the vitreous chamber, which 
lies behind the crystalline lens and in front of the retina and is 
filled with a jelly-like substance, which not only allows the light 
to pass freely but also gives form to the eye. 

The term posterior chamber, or more properly, posterior 
aqueous chamber, is also given to that part of the aqueous cham- 
ber which lies behind the iris and in front of the crystalline lens. 
The posterior chamber is only a narrow chink between the 
peripheral portion of the iris, the suspensory ligament and the 
ciliary processes, and is filled with aqueous humor. 



274 State Board Examinations 

Describe the mechanism of the contraction and expansion of 
the pupil of the eye. 

The pupil, being simply an aperture in the iris, its mechan- 
ism is supplied by the muscles of the latter, which consists of two 
sets of fibers. 

The sphincter of the pupil, which is a narrow band of circu- 
lar muscular fibers, surrounding the pupil on its posterior surface 
5 mm. wide and supplied by the third nerve through the ophthal- 
mic ganglion. 

The dilator of the pupil, which consists of radiating muscular 
fibers converging from the circumference of the iris toward the 
pupillary margin, where they blend with the circular fibers and is 
supplied by sympathetic fibers from the ophthalmic ganglion. 

The pupil contracts and dilates automatically on exposure 
to or protection from light, the greater part of these movements 
being due to the action of the sphincter muscle, which is the 
stronger of the two. 

What is the canal of Schlemm and where is it located? 

This canal, also called the circulus venosus ciliaris, is a small 
channel running completely around the eye at the sclero-corneal 
juncture. Its outer walls are dense, while the inner are com- 
posed of spongy, reticulated tissue, apparently continuous with 
the inner scleral process and closely united with the posterior 
limiting membrane of the cornea and with the pectinate liga- 
ment of the iris, and the meridonal fibers of the ciliary muscle. 

The character of the canal of Schlemm, whether venous or 
lymphatic, was long a subject of controversy, but it is now re- 
garded as a venous sinus, which, through the spaces of Fontana, 
stands in close relation to the anterior chamber on one hand and 
communicates directly with the anterior ciliary veins on the other. 
Usually the canal of Schlemm contains but little blood, because 
it is only a reserve reservoir for the storage of blood when tem- 
porarily retarded in its escape through the anterior ciliary veins. 
In order to maintain normal conditions of intraocular tension, the 
aqueous humor is constantly passing through the space of Fon- 
tana into the canal of Schlemm and thence into the communicat- 
ing veins. 



Anatomy of Eye 21 S 

What kind of a membrane is the conjunctiva and where is it 
located? 

The conjunctiva is a mucous membrane lining the Hds and 
covering the anterior surface of the eyeball and, therefore, it is 
appropriately divided into the palpebral and ocular conjunctiva. 
The annular fold which marks the limit of the conjunctival sac 
is known as the fornix, which permits of free motion of the ball 
in all directions. 

The ocular conjunctiva is smooth and glistening, while the 
tarsal conjunctiva shows a peculiar velvety appearance. The 
conjunctiva in many places is loosely attached and elastic, allow- 
ing it to be moved readily to and fro and at the same time affords 
an opportunity for the accumulation of extravasated fluids. 



What is the function of the lachrymal gland and where is it 
located? 

The lachrymal gland is located in a depression in the roof 
of the orbit at its upper and outer portion. Its function is to 
secrete a fluid called the tears, which keep the eye moist and lubri- 
cate it. Under the disturbing influence of emotion, a foreign 
body in the eye, and inflammation, an excessive secretion takes 
place, causing a running of the nose and an overflow on the 
cheeks. 

How is the crystalline lens held in position? . ■ ^ 

The position of the crystalline lens is maintained by means 
of a series of delicate bands which pass from the vicinity of the 
ora serrata over the ciliary processes, to be attached to the 
periphery of the lens; these fibers constituting the suspensory 
ligament or zone of Zinn. This structure is of great importance, 
not only for the support of the lens, but also in effecting changes 
in the curvature of the lens in the act of accommodation. The 
suspensory ligament is a delicate annular band about 6 mm. in 
width, which blends with the periphery of the lens on one hand 
and with the hyaloid membrane in the vicinity of the ora serrata 
on the other. 

The zone of Zinn is not a continuous membrane but a series 
of interlacing bands or fibers, which are divided into chief and 



276 State Board Examinations 

accessory. The chief zonular fibers constitute the principal union 
between the lens and the surrounding ciliary body, while the 
accessory fibers comprise numerous shorter bands, which act as 
bracers and binders to the chief fibers, and hence are important 
additions to the strength of the suspensory ligament. 



What are the characteristics of the choroid? 

This is the middle coat of the eyeball and is essentially a 
sheet of vascular connective tissue. It extends from the optic 
nerve entrance to the ora serrata, closely united to the retina, 
whose nutrition it furnishes. Although lining the sclerotic, its 
connection with the outer coat is not so firm. 

The most conspicuous of the large blood channels are the 
four venous trunks the venae vorticosae, which pierce the choroid 
at its equator at equidistant points, toward which the smaller 
veins converge. The choroid is dark in color, fragile and easily 
torn. 

What are the angles of the eyelids called? 

They are called outer and inner canthi, from the Latin word 
meaning corners. The singular noun is canthus. 



How is the eye protected from external injury? 

The eyeball itself is protected from injury by being deeply 
placed in the orbit, the edges of which are dense and strong, 
particularly the upper one, which overhangs the eye and is 
capable of shielding it from a powerful blow, as is shown in the 
case of a "black eye," where the blackness is not in the eye but 
in the surrounding soft tissues, which are swollen and inflamed 
and filled with blood, while the eye itself peeps through them 
quite unharmed. If the force of the injury is from below the eye 
is not so well protected. 

The contents of the eyeball are protected by the sclerotic, 
which is a dense, strong, fibrous coat; and, in addition, by reflex 
and automatic action, when any foreign body approaches the 
eye the muscles instinctively contract and the eye is immediately 
closed. 



Anatomy of Eye Til 

Name the humors of the eye and state in what respect they 
differ. 

The humors of the eye from before backwards are the 
aqueous, crystalHne and vitreous. 

The aqueous humor has the consistency of water, with an 
index of refraction of 1.333. 

The vitreous humor is thicker, of jelly-Hke nature, resembling- 
the white of a fresh Qgg, with an index of refraction about the 
same as the aqueous. 

The crystalHne lens is of still greater consistency than the 
vitreous, with the nucleus denser than the cortex. The index of 
refraction of the crystalline as a whole is about 1.44. 

They differ in their density and refractive power, the 
crystalline leading in each particular. 



Describe the eyelids. 

The lids from without inwards are composed of the following 
structures : 

Skin, thin layer of connective tissue, fibers of the orbicularis 
palpebrarum which closes the lids, small plates of cartilage, known 
as the tarsal cartilages and composed of dense fibrous tissue, and 
giving form and support to the lids. Meibomian glands, whose oily 
secretion prevents adhesion of the lids and mucous membrane 
known as the conjunctiva. 

In addition to the orbicularis muscle, there is the levator 
palpebrae superioris, which raises the upper lid and opens the eye. 

The upper lid is much the largest and thus gives better 
protection to the eyeball. The opening between lids is the 
palpebral fissure or commissure, and the angles formed by the 
lids are the outer and inner canthi. 



Name and briefly describe the coats of the eyeball. 

The external coat is the sclerotic, a tough fibrous coat for 
protection. 

The middle coat is the choroid, the vascular and pigmen- 
tary coat. 

The internal coat is the retina, the nervous coat on which 
the images are formed. 



278 State Board Examinations 

Describe the cornea. Give its radius of curvature and its 
refractive index. 

The cornea is the transparent anterior portion of the external 
coat. It is composed of five layers: 

1. Epithelium. 

2. Anterior limiting membrane, or Membrane of Bowman. 

3. The proper substance of the cornea. 

4. Posterior limiting membrane, or Membrane of Descemet. 

5. Endothelium. 

Its index of refraction is 1.33, and its radius of curvature 
7.8 mm. to 8 mm. 

Enumerate the various external motor muscles of the eyeball, 
and give the action of each. What is meant by conjugate movement 
of the eyes? 

The superior, inferior, external and internal recti, the actions 
of which are to turn the eye up, down, out and in; and the 
superior and inferior oblique, which produce a wheel-like move- 
ment, the former rotating down and out, and the latter up and 
out. 

The associated movements of the eyes are both eyes upward 
or downward, both eyes turn to right or to left, in all four of 
which the axes of the eyes remain parallel; and the two eyes 
turning inward in the act of convergence, where the parallel 
condition of the visual axes is destroyed. 



Name the appendages of the eye. 

The eyebrows, the eyelids, capsule of Tenon, conjunctiva, 
the extra-ocular muscles and the lachrymal apparatus. 



What muscles are supplied by the third nerve, and what is the 
possible result if the impulses go wrong? 

The third cranial nerve supplies the superior, inferior and 
internal recti, the inferior oblique, the ciliary muscle and the 
circular fibers of the iris. If there is any interference with its 
function, the action of one or more or all of these muscles will 
be impaired. 



A natomy of Eye 279 

What are the abducens muscles of the eye, and why so called? 

The external recti which turn the eye outwards; the word 
"abducens" is derived from two Latin words, meaning "to lead 
from." 

What is the nature of the aqueous humor of the eye? 

It is a watery fluid with a sHght quantity of chloride of 
sodium in solution, and with an index of refraction about the 
same as water. If lost by injury or operation, it is quickly re- 
placed. It has free communication with the spaces of Fontana, 
the canal of Schlemm and the lymphatics of the eye. 



Describe the Meibomian glands? 

These are the sebaceous glands of the eyelids, numbering 
from 30 to 40 in the upper lid, and 20 to 30 in the lower. Each 
gland has a duct, these ducts terminating as minute points 
arranged in a row along the margin of the lid near its inner 
border. The oily secretion of these glands hinders the overflow 
of tears and prevents adhesion of the lids. 



As a rule which meridian of the eye has the greater curvature? 

This question should read which meridian of the cornea, 
and the answer would be the vertical. 



Where is the fovea centralis and why is it so important? 

It is a funnel shaped depression near the center of the 
macula latea. It is the region of most acute vision, and because 
this acuity of vision is confined to such a small point, the eye 
must move when scanning a surface of any size. 

Its diameter is from 0.2 to 0.4 mm., and the retina at its 
bottom is thinner than at any other place. On account of this 
thinness of the retina allowing the pigment to show through, 
the fovea appears as a dark brown spot. 



ophthalmoscopy 

Describe the indirect method of ophthalmoscopy, give the 
character of the image seen and state where it is located. 

In the indirect method of the ophthalmoscope, a strong con- 
vex lens is used in connection with the instrument. This lens may 
be + 13 D. or + 16 D. and is held at its focal distance in front 
of the eye. The ophthalmoscope is held at a distance of 12 to 15 
inches from the eye, and the light is reflected through the convex 
lens into the eye. The rays returning from the retina of the eye 
under observation form a real, inverted, aerial image between the 
lens and the ophthalmoscope. If the patient is requested to turn 
his eye in accordance with the observer's wishes, every part of 
the fundus may be brought into view. This aerial image may be 
magnified and hence better seen by rotating a convex lens of 3 
D. or 4 D. into the sight hole of the instrument. 

The indirect method gives a larger field than the direct, but 
the details are less magnified, usually about 5 diameters as com- 
pared with sixteen diameters in the direct. The stronger the con- 
densing lens, the larger the field, but the smaller the details; 
the w^eaker the condensing lens, the smaller the view of the fundus, 
but the larger the details. 



Name the various methods of using the ophthalmoscope. How 
do the images differ? What are the advantages of each method? 

There are two methods of using the ophthalmoscope — the 
direct and the indirect. 

In the direct the image is erect and magnified. In the in- 
direct, inverted and smaller than in the direct. 

The advantages of the direct method are the greater magni- 
fication and the disk seen in the erect position. 

The advantages of the indirect method are the larger field 
of view, can be seen through a smaller pupil and not necessary to 
get so close to objectionable patients. 

280 



ophthalmoscopy 281 

In ivhat manner is the ophthalmoscope or mirror used in cases 
of matured cataract, so as to determine the extent of retinal light 
sensibility in support of a favorable prognosis to operation? 

Before the whole lens has become opaque examination with 
an oblique light will throw a shadow of the iris on the cataractous 
part on the side from which the light comes, which shadow is 
absent when the cataract has become mature. 

If the retinal sensibility is unimpaired, the perception of light 
is never lost even in mature cataract ; hence when light is reflected 
into the eye the patient should be able to quickly recognize the 
direction in which it is moved. 



How are errors of refraction detected by the direct method of 
ophthalmoscopy? 

The accommodation of both optometrist and patient must be 
relaxed and the former must wear his correcting lenses or make 
allow^ance for his error. 

Hypermetropia is detected and measured by the strongest 
convex lens with which the fundus can be seen; myopia by the 
weakest concave lens that makes the fundus clear, and astig- 
matism by the difference in the lenses required for vertical and 
horizontal vessels. 



How is the refraction of the eye measured by the indirect method 
of ophthalmoscopy? 

As the object lens is withdrawn, if the disk remains unchanged 
in size, emmetropia is indicated; if it diminishes in size, hyper- 
metropia; and if it increases in size, myopia. This is a method for 
detection of these errors, but cannot be used for accurate 
measurements. 

How is the examination with the ophthalmoscope made by the 
direct method? 

Looking directly into the eye, examining patient's right eye, 
observer uses his right eye and holds ophthalmoscope in his right 
hand. Commencing the examination at a distance of at least ten 
or tw^elve inches and then without losing the reflex approaching 



282 State Board Examinations 

as close as possible in a slow and easy manner, until the full 
details of the fundus are distinctly visible. 



How is the examination with the ophthalmoscope hy the indirect 
method? 

The observer does not have to get so close to his patient, 
but holds the ophthalmoscope at a distance of twelve inches. A 
strong convex lens is required, which is held at its focal distance 
in front of the eye. The light is reflected through the convex lens 
into the patient's eye and the slight movements made that may 
be necessary to bring the disk and vessels into view. The observer 
is supposed to see a real inverted image formed in the air betw^een 
the lens and the ophthalmoscope, this image being also laterally 
transposed. In order to get a better view of the image the 
operator may use a convex lens (about 4 D.) in the sight hole of 
the ophthalmoscope. 

Explain how the ophthalmoscope may he used to detect and 
measure errors of refraction? 

By means of the direct method w^th the accommodation 
of both patient and observer at rest, the strongest convex lens or 
the weakest concave lens rotated into the sight hole of the instru- 
ment, with which the details of the fundus can be clearly seen, will 
represent the amount of hypermetropia or myopia respectively. 



In what way does the picture of the retina as seen by the direct 
method of ophthalmoscopy differ from that seen with the indirect 
method? 

In the direct method the image is erect and more magnified, 
with a restricted field. In the indirect method the image is 
inverted, less magnified, but with a larger field of view. 



Why is refracting with the ophthalmoscope apt to be incorrect, 
and when will it he the most reliable? 

Because of action of the accommodation in patient or optom- 
etrist. In order to obtain accurate results in measuring the refrac- 



ophthalmoscopy 283 

tion with the ophthalmoscope the accommodation of both ob- 
server and observed must be at rest, which is almost an impossi- 
bility. 

The result of such a test would be most reliable if the accom- 
modation of both was under the influence of a cyclopegic, or if 
both were past the presbyopic age. But as a matter of fact the 
ophthalmoscope has been supplanted by the retinoscope for 
measuring the refraction. 



Where are the images formed in the direct and indirect methods 
of the ophthalmoscope? 

In the direct method the image is virtual, erect and enlarged, 
and appears to be some distance behind the eye. 

In the indirect method the image is real, inverted and smaller; 
it is an aerial image formed between the convex lens and the 
ophthalmoscope. 

Which gives the best view of the fundus of the eye, the direct 
method of ophthalmoscopy or the indirect method? 

The direct method gives an erect magnified view of the fun- 
dus. The indirect affords an inverted less magnified view but with 
a much larger extent of the surface of the fundus visible. On 
account of the magnification the direct method is usually con- 
sidered the best. 

Describe the optic disk as seen by the ophthalmoscope when 
in a normal condition. What peculiarities are sometimes noted in 
the fundus of the normal eye? 

The optic disk is the most conspicuous landmark of the 
fundus and is the head of the optic nerve. It is nearly circular or 
slightly oval, with a diameter of 1.5 mm., which is magnified 
fifteen times by the direct method of the ophthalmoscope. The 
disk projects slightly above the level of the fundus and presents 
a central depression. The color is whitish or pinkish, and 
presents a marked contrast with the red of the fundus. The 
edges of the disk should be clearly defined. 

There are variations in the color of the fundus, shape of the 
disk, width of the scleral and choroidal rings, distinctness of the 



284 State Board Examinations 

margins, color and size of the blood vessels, pulsation of retinal 
veins, etc. 

In direct ophthalmoscopy an optometrist whose correction is 
-{-2D. sphere combined with a — 1 cylinder axis 180° finds that 
with naked eye he can see the blood vessels in the disk in the horizontal 
meridian- with a -\- 1 D. lens in the ophthalmoscope, hut requires 
-j- 2 D. lens to see the blood vessels in the vertical meridian. Write 
a prescription for the eye being examined. 

An analysis of this sphero-cylinder will show that the 
optometrist has compound hypermetropic astigmatism, his 
correction being + 1 D. vertically and + 2 D. horizontally. In 
the first place it must be remembered that he sees the horizontal 
blood vessels with the vertical meridian of his eye, and the vertical 
blood vessels with the horizontal meridian of his eye. Therefore, 
if he requires + 1 D. to see the horizontal vessels that means 
+ 1 D. for his vertical meridian, and if he requires + 2 D. for 
the vertical vessels, that means + 2 D. for his horizontal meridian. 

As these are exactly the powers required in each meridian 
to correct his own astigmatism the refraction of the patient is 
proven to be emmetropic and no glasses are required. 



What is the principal optical point in the use of the ophthalmo- 
scope by the direct method? 

That the rays emerging from the observed eye may be 
exactly focused upon the retina of the observer, and these emer- 
gent rays must be of such character, either naturally or made so 
artificially, that this may be accomplished with the accommoda- 
tion of both parties at rest. 

If the observer is emmetropic and the patient emmetropic 
also, the emerging rays will be parallel, and will focus on retina 
of observer without accommodative effort. If the patient is 
hypermetropic the emerging rays will be divergent and a convex 
lens will be required to focus them in observer's eye, presuming 
the accommodation of both to be passive. 

If the patient is myopic the emerging rays will be convergent 
and a concave lens will be required for the observer to get a 
good view of the fundus. 



ophthalmoscopy 285 

In measuring ametropia by the ophthalmoscope, what must be 
allowed for and what function must be passive? 

Any error of refraction in the eye of the optometrist must be 
allowed for unless he wears his correcting lenses, and the accom- 
modation of both patient and observer must be at rest. 



What is the main purpose of the ophthalmoscope? 

To determine the transparency of the refracting media, to 
examine the fundus and note the condition of the optic disk 
and macula; in short to ascertain the absence or presence of 
disease, and in the case of the latter its character and location. 



In the direct method of the ophthalmoscope the examiner finds 
that with the naked eye he can see the blood vessels of the disk in 
the horizontal meridian with a -\- 1 D. lens in the instrument, but 
requires a -\- 2 D. lens to see the blood vessels in the vertical meridian. 
The error of refraction in the examijiers eye is corrected by -\- 2 D. S. 
^ — 1 D. cyl. axis 180°; what is the correction for the eye under 
the examination? 

In the estimation of the refraction of an astigmatic eye by 
the direct method of the ophthalmoscope, it must be remembered 
that when looking at a vertical blood vessel the observer sees it 
through the horizontal meridian, and when viewing a horizontal 
vessel, through the vertical meridian. 

The wording of this question is ambiguous, as for instance, 
the statement "the blood vessels of the disk in the horizontal 
meridian." We do not know if the framer of the question means 
the vertical vessels which must be seen through the horizontal 
meridian or the horizontal vessels which would be seen through 
the vertical meridian. 

Judging from the correcting lenses of the examiner, we know 
his vertical meridian is hypermetropic 1 D., and his horizontal 
meridian hypermetropic 2D., and as. he is making the examina- 
tion with his naked eye, this error of refraction must be allowed 
for. If the "vessels in the horizontal meridian" means vertical 
vessels seen through the horizontal meridian, for which the 
correction is + 1 D., while the refraction of this meridian in 



286 State Board Examinations 

the examiner's eye is hypermetropic 2 D., we must assume the 
patient is myopic 1 D. in this meridian. 

And from a Hke point of view, if + 2 D. is the correction in 
the sight hole of the instrument for the vertical meridian, while 
the examiner is hypermetropic only to the extent of 1 D. in this 
meridian, then the presumption is that the patient is also hyper- 
metropic 1 D. in this meridian. 

The patient's correction for the two meridians would be 
- 1 D. cyl. axis 90° C + 1 D. cyl. axis 180° or - 1 D. C + 2 
D. cyl. axis 180°. 

But if the words "vessels in horizontal meridian" mean 
horizontal vessels, then the + 1 D- required is the correction for 
the vertical meridian. And as this + 1 D. is required to correct 
the examiner's hypermetropia in this meridian, we must assume 
the patient's refraction is emmetropic in this meridian. 

And in like view of the case the + 2 D. required for the 
vertical vessels, would indicate the correction for the horizontal 
meridian. And as a lens of like amount is required to correct 
the examiner's hypermetropia in this meridian, the inference is 
the patient is emmetropic in this meridian. 

From the latter interpretation of the case, we would have no 
prescription to write, as the patient w^ould be shown to be emme- 
tropic. 

What are the optical principles of ophthalmoscopy in emme- 
tropia by the direct method? 

In ophthalmoscopy when the interior of the eye is illuminated 
by the mirror the fundus becomes a luminous body and gives off 
rays of light. As these rays pass out of the pupil they are subject 
to refraction of the dioptric media of the eye, and can be used 
by an observing eye so as to have an image of this fundus formed 
upon its retina. It was the theory of Helmholtz, who gave us 
ophthalmoscopy, to bring the illuminated fundus of the observed 
eye and the retina of the observing eye in positions of conjugate 
foci. 

He reasoned that if the fundus of an eye became the source 
of light, the rays in passing out through the optical system of 
the eye w^ould be refracted and proceed toward the conjugate 
focus of the object from which they came. If now the eye of 



ophthalmoscopy 287 

an observer could be placed in the path of these emergent rays 
so that they could be focused on its retina, the fundus from which 
these rays proceeded could be seen in all its details. 

In emmetropia the rays will emerge from the observed eye 
parallel. If now the eye of the observer that is placed in the 
path of these returning rays is also emmetropic and as such 
adapted for parallel rays, the retina of the observed eye and of 
the observer's eye will be at conjugate foci, and an image of the 
fundus of the first will be formed on the retina of the second. 
Therefore an emmetropic eye can see clearly the details of the 
fundus of another emmetropic eye when both eyes are in a static 
condition and without the intervention of any lens. 



What are the optical principles of ophthalmoscopy in myopia 
by the direct method? 

In myopia the rays as they emerge are convergent toward 
the far point situated at some finite distance. The eye of the 
observer if emmetropic and as such adapted for parallel rays, 
could not focus these convergent rays on its retina. 

In order to accomplish this, the rays must first be made 
parallel, which can be done by the interposition of a concave 
lens of such strength as will lessen the convergence of the rays 
and make them parallel. The negative focus of this lens will 
coincide w^ith the location of the far point and the amount of 
myopia of the observed eye. 

In this way the opthalmoscope may be used in the direct 
method to measure the degree of myopia. When the emmetrope 
looks into the eye of a myope, the fundus is very much blurred. 
A concave lens rotated into the sight hole of the instrument 
clears it up and the weakest lens that affords the clearest view of 
the optic disk and vessels will be the measure of the myopia; 
such lens making the emergent convergent rays parallel and 
bringing the retinae of the two eyes into positions of conjugate foci. 



What are the optical principles of ophthalmoscopy in hyperme- 
tropia by the direct method? 

In hypermetropia the far point is negative and the rays as 
they emerge from the observed eye are divergent. The eye of 



288 State Board Examinations 

the observer if emmetropic and in a static condition, when placed 
in the path of these divergent rays, is unable to focus them upon 
its retina until they have been rendered parallel. 

This can be accomplished by the interposition of a convex 
lens, whose focus would correspond to the negative focus back 
of the retina. The two retinae are then placed in the positions 
of conjugate foci and the image of the fundus of the observed 
eye is formed on the retina of the observing eye. 

A convex lens is rotated in the sight hole of the instrument 
and the strongest convex lens with which the clearest view of 
the fundus can be obtained will be the measure of the hyperme- 
tropia, such lens making the emerging divergent rays parallel 
and bringing the retinae of the two eyes into positions of con- 
jugate foci. 

What are the optical principles of ophthalmoscopy hy the 
indirect method? 

The convergent rays that emerge from a myopic eye form 
an inverted image of the fundus of such eye at the far point, 
and if the observer places himself at the proper distance so that 
his own far point coincides with this image he will be able to 
see it distinctly. 

In the indirect method of the ophthalmoscope we make the 
observed eye artificially myopic by means of a strong convex 
lens placed in front of it. It is customary to use a + 20 D. 
lens, which in the case of an emmetropic eye would form an 
inverted image of the fundus 2 inches in front of the lens, which 
would then be its conjugate focus. 

An emmetropic observer at the usual reading distance by the 
aid of his accommodation, would be able to bring his far point 
to the position of this image, which would then be pictured on 
his retina. 

If the eye under examination be myopic the aerial image 
will be a little closer, and if the eye be hypermetropic a little 
farther, than in emmetropia. 

Should the observer be presbyopic and unable to use his 
accommodation, a convex lens rotated into the sight hole of the 
ophthalmoscope whose focal length would coincide with the 
aerial image would render the rays coming from it parallel, so 
that they could be focused by the eye of the observer. 



ophthalmoscopy 289 

If the observer is hypermetropic a stronger lens will be 
necessary; if myopic a weaker lens or no lens at all if the myopia 
was of such amount that its far point coincided with the aerial 
image. 



Retinoscopy 

Where is the nodal point of the observing eye with respect to the 
observed fundus located when the point of reversal in skiametry has 
been reached? 

Conjugate to it. 



Describe the behavior of the plane and of the concave mirror in 
skiametry. 

The rays of light proceeding from the round opening in the 
chimney are divergent, and after striking the plane mirror are 
reflected in a continuance of the divergence, just as if they came 
from a point as far back of the mirror as the chimney is in front 
of it; hence, the shadow moves with in hypermetropia and emme- 
tropia, and against in myopia. 

With the concave mirror, on the other hand, the rays which 
leave the source of light as divergent, are reflected from the 
mirror convergently and brought to a focus and cross before 
entering the observed eye, and consequently the shadow moves 
against in emmetropia and hypermetropia, and with in myopia. 

In both cases we proceed to estimate the refraction in the 
same way, that is, by neutralization by the indicated lenses. 



What is indicated by the form of the shadow in skiametry? 
by the direction of the shadow? by the speed of the shadow? 

1. The form of the shadow indicates the character of the 
error, whether astigmatic or spherical. If the shadow is curved 
or crescent-shaped at its edge, the indications point to simple 
myopia or hypermetropia. If the edge is straight or shows a 
banded appearance, the indications point to astigmatism. 

2. The recognition of the direction of the shadow is a most 
important point, as it indicates the character of the refraction. 

Using the plane mirror, if the direction is "with," it may be 
emmetropia or myopia less than 1 D., but we usually suspect 

290 



Retinoscopy 291 

hypermetropia, which is proven to be present if the movement 
continues "with" after a + 1 D- lens is used to neutralize the one 
meter distance. If the direction is "opposite," the indications 
are for myopia of more than 1 D. 

3. The speed of the shadow is affected by the movement of 
the mirror itself but making due allowance, the speed indicates 
if the error is high or low. When the refractive error is high, there 
is a slowness in the rate of movement of the shadow calling for a 
strong lens for its neutralization; whereas, when the movement 
seems to be fast, the error is likely to be low and call for a weak 
lens. 



Where is the ohservijig eye with the concave mirror located when 
the shadow movement is with the 7nirror? 

Beyond the point of reversal. 



How is the shadow movement in skiametry affected by low 
degrees of ametropia? Give the reason. 

In low forms of ametropia the movement is fast because of the 
nearness of the point of reversal. The nearer to the point of re- 
versal the faster the movement, because at this point the emergent 
rays from the observed eye are conjugate to the observing eye. 



Give and explain the reason for a straight-lined contour of the 
shadow in skiametry. 

A straight edge of the shadow indicates astigmatism, and is 
due to the fact that the emergent rays from the observed eye are 
refracted by the two chief meridians of the eye, one of least and 
one of greatest curvature, and as a result the focus could never 
be round or a point, but always oval or a line. 



What kind of lens should be placed before the observed eye to 
produce neutralization when the observer with the concave mirror 
finds a shadow movement against the mirror? 

Convex. 



292 State Board Examinations 

With respect to the degree of ametropia what may be deduced 
from a bright retinal reflex in skiametry? 

In high errors the reflex is dull; in low errors it is bright. 



In what respect does the dynamic method of skiametry differ 
from the static method? 

In static skiametry the test is made with the ciliary muscle 
at rest. For this purpose a cycloplegic is sometimes used by 
medical men, while optometrists endeavor to get the same result 
by darkening the room and asking the patient to look carelessly 
at some distant object. 

Dynamic skiametry is the opposite of static, and signifies the 
use of some force, which in the case of the e3^e is known as the 
accommodation. The test is made while the accommodation is 
in action, which is done by asking the patient to read test letters 
placed on the examiner's brow or attached to his mirror. 



In static skiametry with the plane mirror, the operator working 
at one meter, no movement of the shadow is found in the vertical 
meridian; with a 1 D. spheric lens placed before the eye there is 
no movement of the shadow in the horizontal meridian. State the 
nature and the amount of the ametropia. What was the direction 
of the shadow movement when the mirror was rotated horizontally 
before the imposition of the lens? 

If with plane mirror at one meter no movement is found in 
vertical meridian, the refraction of this meridian is myopic 1 D. 
If with + 1 D. lens there is no movement in the horizontal 
meridian, this meridian is em.metropic. The case then is one of 
simple myopic astigmatism, and would be corrected by a — 1 D. 
cyl. axis 180°. 

The direction of the shadow in the horizontal meridian be- 
fore the use of the convex lens, would be "with." 



In skiascopy ivhat two poi7its are conjugate foci? 
The retina and the point of reversal. 



Retinoscopy 293 

With the plane mirror, what is the direction of the shadow 
movement when the observer is within the point of reversal^ 

With. 



I7i dynamic skiametry, with fixation at 40 inches, the point 
of reversal in the vertical meridian is found at 26 inches, + 2.50 D. 
lens being before the observed eye and at 20 inches in the horizontal 
meridian with — 1.25 D. lens before the observed eye. What is the 
kind and amount of ametropia? 

Mixed astigmatism. Vertical meridian hypermetropic 2 D. 
and horizontal meridian myopic 2.25 D. 

Correcting lens: + 2 D. cyl. axis 180° O — 2.25 D. cyl. 
axis 90°. 



Put the following back to the retinoscope findings after reversing 
the shadow from 40 inches: — .50 ^ — .25 X 50. 

If we reduce this sphero-cylinder to a cross cylinder, we have 
— .50 D. cyl. axis 1.50 O — .75 D. cyl. axis 60° 
which would indicate a myopia of .50 D. in the 60th meridian and 
of .75 D. in the 150th meridian. 

Inasmuch as when working at 40 inches we must add — ID. 
to the retinoscopic findings, therefore the latter must have been 
+ .50 D. in the 60th meridian 
+ .25 D. in the 150th meridian. 



Dark room findings: 

0. D. + 1.50 axis 90° O + 3 axis 180° 
O. S. + 2.25 axis 90° C + 1-75 axis 180° 

Write the prescription. 

If we tak€ this literally and make allowance for a working 
distance of 1 D., the prescription would be 

O. D. + .50 D. sph. C + 1.50 D. cyl. axis 180° 
O. S. + .75 D. sph. C + .50 D. cyl. axis 90° 
But usually in retinoscopy we find the lens that corrects 
each meridian and then we have to deal with meridians instead 
of as above with axes. 



294 State Board Examinations 

Usually the dark room findings come to us in this shape: 
O. D. + 1.50 vertically — + 3 horizontally, 
O. S. +2.25 vertically — + 1.75 horizontally, 
which means that the lenses mentioned neutralize the movement 
in the meridians mentioned. Then making the allowance of 1 D. 
and remembering that axes are at right angles to meridians, the 
prescription would read : 

O. D. + .50 D. sph. C +1.50 D. cyl. axis 90° 
O. S. + .75 D. sph. C + .50 D. cyl. axis 180° 



What is meant by the statement in skiascopy that the shadow 
moves with the mirror, or the shadow moves against the mirror? 

When the patient's pupil is properly illuminated the observer 
will see a bright reflex from the pupillary area. As the mirror 
is turned the illumination moves off followed by the shadow, 
thus causing the movement of the shadow. 

The illumination on the patient's face always moves in the 
same direction as the mirror is rotated, but not necessarily so in 
the pupil, where it sometimes appears to move opposite, and 
hence we speak of the shadow moving ''with" or "against" the 
movement of the mirror. 

What characteristic conditions exist when the point of reversal 
is reached? 

At the point of reversal no definite movement of the retinal 
illumination can be made out, and while the pupil is uniformly 
illuminated it shows a duller reflex than when within or beyond 
this point, and the shadow will be absent. 



What difference is observed in the directions of the movement 
of the facial and retinal illuminations by the plane and the concave 
mirror respectively? 

The movement on the face is always the same as the mirror, 
whether it be convex or concave. 

In the pupil the movements caused by the plane and the 
concave mirror are the reverse of each other, and they also vary 
with the condition of refraction. 



Retinoscopy 295 

In emmetropia, hypermetropia and myopia less than 1 D. 
the movement is -with when using a plane mirror and against 
when using a concave mirror. 

In myopia greater than 1 D. the movement is against with 
a plane mirror and ivith when using a concave. 



With respect to the point of reversal, where must the observer 
he located to see (a) the shadow movement directed the same as the 
plane mirror; (b) the shadow movement directed opposite to the 
plane mirror; {c) the shadow movement directed the same as the 
concave mirror; (d) the shadow movement directed opposite to the 
concave mirror? 

a. Inside the point of reversal. 

b. Outside the point of reversal. 

c. Outside the point of reversal. 

d. Inside the point of reversal. 



What may be discovered by dynamic skiametry? In what 
respect does the dynamic method differ from the static method? 

The object of dynamic method is to discover spasm of the 
accommodation. It differs from the static method in that a 
definite amount of accommodation is brought into action accord- 
ing to the distance of the point of fixation, whereas in the static 
method the effort is to keep the accommodation at rest. 



Which focal point is sought to be determined by skiametry and 
what relation exists between it and the retina of the observed eye? 

The anterior focal point which would be conjugate to the 
retina of the eye under observation. 



Describe the respective shadow movement with the plane and 
concave mirror, and state the advantage, if any, of the former over 
the latter? 

With the plane mirror the shadow movements are in the 
same direction as the movements on the face when the eye 



296 State Board Examinations 

cf observer is inside the point of reversal, and in the opposite 
direction to the movements on the face when the observer's eye 
is beyond the point of reversal. 

The movements caused by a concave mirror are just the 
reverse in each case to those caused by a plane mirror. 

The advantages of the plane mirror are that it is simpler 
and easier and can be used at any distance, whereas with the 
concave mirror the eye under observation must be at such a 
distance that the rays of light from the mirror come to a focus 
and cross before entering it. 



Hoiv does the observer decide that the point of reversal has been 
reached? 

The point of reversal is reached when the movements have 
been neutralized, or when it is impossible to determine in which 
direction the shadow appears to move. This is not always an 
easy matter, and the ability to quickly find the point of reversal 
comes only after extended practice. 

Perhaps the better way for a beginner is to make or estimate 
between the lens that just makes the movement with and one 
that just turns it against. For instance, if with + 2 D. the move- 
ment is still with and with + 2.50 the movement has turned 
against, we have narrowed the point of reversal down to a point 
between the strength of these two lenses. 



Describe the appearance of the retinal illiimijiation in pro- 
nounced astigmatism. 

As a band of light which is characteristic of astigmatism. 
If the spherical error be high and the astigmatism slight, the band 
of light will not be noticeable on first inspection, not until the 
axial error has been at least partly corrected, and becomes 
brio;htest when fuUv corrected. 



Describe the directions of the real movement of the facial and 
retinal illuminations by the plane and the co72cave mirror respec- 
tively. 



Retinoscopy 297 

The facial illumination is always in the same direction as 
the movement of the mirror, whether it be plane or concave. 

With a plane mirror the retinal illumination is always in 
the same direction as the movement of the mirror, whether the 
case be one of emmetropia, hypermetropia or myopia. The 
concave mirror causes reverse m.ovements. 



Describe the directions of the apparent movement of the facial 
and retinal illuminations by the plane and the concave mirror 
respectively. 

The apparent moA'ement of the retinal illumination depends 
upon whether the observer is within or beyond the point of 
reversal, or the point conjugate to the retina of the observed eye. 

Using a plane mirror if the observer is within the point of 
reversal (as in emmetropia and hypermetropia) the movements 
will be with. If observer is beyond the point of reversal (as in 
myopia greater than ID.) the movements will be against. 

Using a concave mirror, if the observer is within the con- 
jugate point, the movements will be against; if beyond the 
conjugate point, with. 

Explain the dynamic method of skiametry. State the purpose 
of this method. 

In the dynamic method of skiametry a card of letters is 
attached to the retinoscope or placed on the brow of the examiner, 
at which the patient is asked to look, after which lenses are used 
in the customary way to find the point of reversal. 

In this method a definite amount of accommodation is used, 
depending upon the distance of fixation, in this way reducing the 
tendency to spasm. The test can be made at any distance, and 
the result represents the amount of defect, no allowance to be 
made for the working distance. 



// with the plane mirror before the static eye, a shadow move- 
ment against the mirror is neutralized in the horizontal meridian 
by — 1 D. lens of 40 inches and in the vertical meridian by the 
same lens at 30 ijiches (a) what will be the movement at 20 inches? 
(b) what is the amount of ametropia? 



298 State Board Examinations 

As the mirror must be moved up to 30 inches in order to 
neutraUze the vertical meridian, this meridian must be .?)Z D. 
more myopic. As the rule is to add — 1 D. to the retinoscopic 
findings, we have — 2 D. as the measure of the horizontal 
meridian, and — 2.33 D. as the measure of the vertical meridian. 
This is a case of compound myopic astigmatism, and the correct- 
ing lens would be 

-2D. sph. C - .^^ D. cyl. axis 180° 

At 20 inches there would be no movement in the horizontal 
meridian because neutralized by the 2 D. of myopia in this 
meridian, while in the vertical meridian there would be a move- 
ment against. 

With a -\- 1 D. lens before the static eye and the plane mirror 
at 1 meter the shadow movement is with the mirror in the horizontal 
and against it in the vertical meridian; it is also found that there 
are other superposed lenses of 1 D., each of which separately neutral- 
izes these movements. Give the character of the ametropia and of 
the lens that corrects it. 

If with + 1 D. lens before the static eye, to neutralize the 
distance, the movement is with in the horizontal meridian, hyper- 
metropia is proven, and if this movement is neutralized by 
another 1 D. lens, it must be a + 1 D., which would be the 
measure of this meridian. 

If in the vertical meridian the movement is against myopia 
is proven, and if this movement is neutralized by a 1 D. lens 
it must be a — 1 D. lens, which would be the measure of this 
meridian. 

The case is therefore one of mixed astigmatism and the 
correcting cross-cylinder would be 

+ 1 D. cyl. axis 90° C - 1 D. cyl. axis 180° 
which can be transposed to 

- 1 D. sph. C + 2 D. cyl. axis 90° 



A patient, while wearing — 2D. sph. lenses under the dynamic 
test, neutralizes the shadow movement during fixation up to 40 cm. 
Give his amplitude of accommodation and the distance of his near 
point without glasses. 



Retifioscopy 299 

In order to neutralize the movement at 40 cm. 2.50 D. of 
accommodation is necessary, and as a further 2 D. of accom- 
modation must be used to overcome the — 2D. lenses worn, 
the amplitude of accommodation must be 4.50 D., which would 
represent a near point of 9 inches. 



A young hyperope, having normal acuteness of vision, either 
with or without + 0.5 D. lenses, being subjected to the dynamic 
test, is found to require -\- 1 D. lenses to arrest motion of the shadow 
for all proximate distances up to 25 cm. Which lenses should be 
given him for reading? 

The + .50 D. lenses represent the amount of manifest 
hypermetropia as measured by the trial case, and + 1 D. the 
amount of total error as measured by the dynamic method of 
retinoscopy. 

Theoretically it would be proper to give him the full correc- 
tion of + 1 D. for reading, but practically it is sometimes advis- 
able in young people to slightly undercorrect. 



By static retinoscopy at 40 inches the vertical meridian of the 
eye is found to neutralize with a + 1.75, and the horizontal meridian 
with a + 1.50. What is the correction for distance? 

After making allowance for the working distance of 40 
inches the refraction of the vertical meridian is hypermetropic 
.75 D. and of the horizontal meridian .50 D. The correcting 
lens would be 

-f .50 D. sph. C + .25 D. cyl. axis 180° 



// a person has an error of — 1 on -\- 2 cyl. axis 45°, what 
movements of shadows will be observed with the plane retinoscope? 

Assuming the test is made at a distance of 40 inches, then 
in the meridian of the axis of the cylinder there would be no 
motion on account of the 1 D. of myopia, while in the 135th 
meridian the motion would be with the mirror. 



300 State Board Examinations 

In using the retinoscope with plane mirror, what causes the 
shadow, and why in hyperopia does the shadow move with the 
mirror? 

In retinoscopy light is reflected by means of a mirror into 
the patient's eye where it falls upon the retina, making an area 
of light, which constitutes the retinal illumination. The shadow 
is the non-illuminated portion of the retina immediately sur- 
rounding the illumination. 

In hypermetropia the rays emerge divergently, consequently 
the focus of such an eye is beyond the observer, in fact it is even 
beyond infinity; therefore, according to the laws of optics under 
such conditions, the shadow must move with the mirror. 



In static skiametry, if the patient has 1 D. of myopic astig- 
matism in vertical meridian, at what distance will the neutral point 
he located in the horizontal meridian with a -\- 1 D. lens placed 
before the eye? 

The patient has 1 D. of myopic astigmatism in the vertical 
meridian w^hile the horizontal meridian is emmetropic. A + 1 D. 
lens placed before the eye would locate the neutral point of the 
latter at one meter, which is just the position of the retina of the 
observer. This is just the same as in emmetropia where + 1 D. 
lens is necessarv to neutralize the movements at one meter. 



With a -\- 3 D. lens before the static eye, where will the neutral 
point in the horizontal and vertical meridians be located for the 
ametrope whose correction is + ID. sph. O -{- 1 D. cyl. axis 90°? 

In this case the correcting combination would show that 
there was + 1 D. of hypermetropia in the vertical meridian, 
and 2 D. of hypermetropia in the horizontal meridian. Placing 
a + 3 D. lens before such an eye would make the vertical me- 
ridian myopic to the extent of 2 D. which would cause the neutral 
point to be located at 20 inches. The + 3 D. lens would make 
the horizontal meridian myopic to the extent of 1 D. thus causing 
its neutral point to be located at 40 inches. 



Retinoscopy 301 

What is the fundamental principle of the test with the skiascope? 

The principle of retinoscopy Is to find the point of reversal 
or the far point which either occurs naturally as In myopia or 
Is artificially produced by convex lenses. 



Which focal point is sought in the examination with the retino- 
scope, and what is the relation of that point with the retina of the eye? 

The point of reversal where no motion Is evident. This is 
the far point of the eye, naturally so in myopia and artificially 
produced by convex lenses in hypermetropla, and is conjugate to 
the retina. 

In static skiametry with the plane mirror the operator, ivorking 
at one meter, no 7novement of the shadow is found in the vertical 
meridian; with a -\- 1 D. sphere placed before the eye there is no 
movement of the shadow in ihe horizontal meridian. State the 
nature and the amount of the ametropia. What was the direction 
of the shadow movement when the mirror was rotated horizontally 
before the imposition of the lens? 

Under the conditions named no movement of the shadow 
would show a myopia of 1 D. in the vertical, and a neutralization 
of the movement by a + 1 D. emmetropia of the horizontal 
meridian. This would indicate a myopia of 1 D. with the rule 
and would be corrected by — ID. cyl. axis 180°. 

The direction of the shadow movement In the horizontal 
meridian before the imposition of the + 1 D. lens must have 
been with the light. 

In static skiametry if the patiejtt has 1 D. of myopic astigma- 
tism, correcting lens axis vertical, at what distance will the point 
of reversal be located in the horizontal meridian with -\- 1 D. placed 
before the eye? 

If the axis of the correcting cylinder Is vertical the defective 
meridian is horizontal, being myopic to the extent of 1 D. A + 
1 D. placed before the eye will make this horizontal meridian 
myopic that much more, and the point of reversal will then be 
located at 20 inches. 



302 State Board Examinations 

Since the layers that make up the retina of the eye are trans- 
parent how is it possible that we can see a reflex in retinoscopy? 

Inasmuch as the retina has a pigment layer and besides is 
in close contact with the dark-colored choroid the light that is 
thrown into the eye is reflected from the retina and shows in 
the pupil as a bright-colored reflex. 



Why do we seek in skiascopy to get neutrality of motion of the 
light reflex? 

In order to find the point of reversal, which is the principle 
of retinoscopy, and by which the refraction of the eye is deter- 
mined. 

Why is it in skiascopy that it is not always possible to flnd 
neutrality of motion of the light reflex? 

On account of irregularity of curvature of the crystalline 
lens or cornea. 

What is meant in skiascopy by the term ^'with the mirror?'' 

That the movement of the retinal reflex in the pupil is in 
the same direction as the movement of the light on the face. 



What is the appearance of the light in the pupil when there is 
astigmatism? 

As a band of light. 

Working with the plane mirror the vertical meridian is neutral 
at one meter and the horizontal meridian at half meter. What will 
be the movements in the two meridians at half meter? 

The vertical meridian is shown to be myopic 1 D. and the 
horizontal meridian myopic 2D. At half meter the movements 
in the vertical meridian would be with, while the horizontal 
meridian will be neutral. 

In a certain eye the motion of one principal meridian is slightly 
against, while that of the other is slightly with. What will be the 



Retinoscopy 303 

effect of putting in front of the eye a high-power plus lens? Also the 
effect of using a high-power minus lens? 

This would be a mild case of mixed astigmatism. A strong 
convex lens would make both meridians myopic, and the motion 
would be against in both. A strong concave lens would make 
both meridians hypermetropic, and the motion would be with 
in both, presuming a plane mirror is used. 



A patient who is 1 D. hypermetropic and whose near point is 
at 13 inches requires a -\- 2 D. lens to show neutral motion hy the 
dynamic method at 20 inches. What conclusions would you draw? 

That there was a hypermetropia of at least 2 D. 



What is the difference between a static test and a dynamic test? 

In a static test the ciliary muscle is supposed to be passive 
or at rest, as in the objective tests. In a dynamic test the ciliary 
muscle is an action, as in the subjective tests. 



A child shows hy the static test with the skiascope an error 
of + 50 D. and hy the dynamic test made at 16 inches he shows an 
error of + 1.25 D. What glasses would you recommend, and why? 

The dynamic test shows that with a relaxation of accom- 
modation the error is equal at least to 1.25 D. We hardly think 
it advisable, however, to prescribe so full a correction on account 
of the strong accommodation of youth. We would be inclined 
to make a compromise and order about + .75 D., if it seemed 
necessary to order glasses at all. 



A patient sixty years of age shows hy the static method with 
the skiascope 2 D. of hypermetropia and hy the dynamic method 
at 20 inches a + 4 D. is required to neutralize motion. What is 
the prescription for distance and for reading at 16 inches? 

+ 2 D. for distance, + 4.50 D. for reading. 



304 State Board Examinations 

Why are the movements of the edge of the light reflex in skia- 
scopy in reversed directions, according as the plane or concave 
mirror is used? 

Because in the case of the concave mirror the rays of light 
have been brought to a focus and diverged. 



What is the difference in the real movement of the retinal 
illumination in skiascopy and the apparent movement? 

The real movement corresponds to the movement of the 
retinoscope, while the apparent movement is dependent on the 
location of the focus, whether in front of or behind the instru- 
ment. 

At what point is the astigmatic hand the most noticeable? 

The band of light is seen when one meridian is corrected 
and the meridian at right angles remains uncorrected or only 
partly corrected. 

When the astigmatism is relatively high and the spherical 
error of small amount the band of light is usually evident on first 
inspection before any neutralizing lens is placed in position. 



When does the edge of light reflex move the most rapidly, in 
high or in low errors? 

When the refractive error is slight and a weak lens is required 
for its correction the movement is fast. 



Upon what principle or principles is rhinoscopy based? What 
is the scissors movement? 

The principle of retinoscopy is the finding of the point of 
reversal, or the far point of a myopic eye, either naturally myopic 
or made so artifically b}^ means of convex lenses that will bring 
the emergent rays of light to a focus at some definite distance. 

In the so-called scissors movement there are two areas of 
light, or two shadows seen in the same meridian, occurring usually 
in astigmatism, and especially in irregular astigmatism. The 



Retinoscopy 305 

cause of the scissors movement is mostly attributed to a slight 
tilting of the crystalline lens. 



In using the retinoscope what glasses would you wear in order 
to get the best results, and why is it necessary to wear glasses at all? 

The reason why it is necessary for the observer to wear 
glasses, and the only reason, is that he may get the clearest view 
of the patient's pupil. For this purpose, if the observer have 
sufficient amplitude of accommodation, his distance glasses will 
answer until an age is reached when there is but little accom- 
modation left, and then a convex lens must be added that will 
afford sharp vision at 40 inches. 



In skiametry why is the point of reversal conjugate to the 
retina of the observed eye? 

Because it is the neutral point, and the rays diverging from 
which would exactly focus on the retina of the eye. 



Why is the plane mirror generally given preference over the 
concave mirror? 

With the plane mirror the test can be made at any distance, 
whereas with the concave mirror care must be taken to see that 
the distance be greater than the focus of the mirror. 



What is the difference of movement of the light edge with plane 
mirror as compared with the concave mirror? 

Just the reverse. In emmetropla and hypermetropia where 
the plane mirror shows a with movement the concave mirror 
will show an against movement. In myopia where the plane 
mirror shows an against movement the concave mirror will show 
a with movement. 

When is the retinal reflex the brightest with the plane mirror, 
and when with the concave mirror? 

With both mirrors close to the point of reversal. 



306 State Board Examinations 

When does the light spot on the retina move the fastest, when 
the reflex is near or far from the neutral point? 

Near the neutral point. 



What are the characteristics of the light reflex at the reversal 
point in the human eye? 

No definite movement of the retinal illumination can be 
made out, and the pupillary area is seen to be uniformly illumi- 
nated. 

On what does the real movement of the retinal reflex depend? 

On the direction in which the light is made to go by the 
tilting of the mirror. 

On what does the apparent movement of the retinal reflex 
depend? 

On the character of the emergent rays. 



When the observer is nearer than the neutral point, what will 
he the direction of the shadow movement with the plane mirror? 

With. 



When the observer is farther away then the neutral point, what 
will be the direction of the movement with the concave mirror? 

With. 

In retinoscopy by the usual static method, if the patient has 
2 D. of hypermetropic astigmatism, axis vertical, at what distance 
will the point of reversal be located in the horizontal meridian with 
a -\- 3 D. sphere placed before the eye? 

If the axis of the correcting cylinder is vertical, the 2 D. 
of hypermetropic astigmatism would exist in the horizontal 
meridian. A -f 3 D. sphere placed before the eye would over- 
correct this meridian and make it myopic to the extent of 1 D., 
in which case the point of reversal would be located at one 
meter or 40 inches. 



Retinoscopy 307 

In retinoscopy by the usual static method, liith a -\- 2 D. lens 
before the eye, iihere uill the point of reversal in the vertical and 
horizontal meridians he located for the ametrope whose correction is 
-2D. sph. -ID. cyl. axis 180°? 

This is a case of compound myopic astigmatism in which 
naturally the point of reversal for the vertical meridian would 
be located at 13 inches and for the horizontal meridian at 20 
inches. The addition of a + 2 D. before the eye would increase 
the myopia to that extent, and then the point of reversal for the 
vertical meridian would be moved up 8 inches and for the hori- 
zontal meridian to 10 inches. 



Using the method of dynamic skiametry and working at 40 
inches, a -\- 1 D. sphere causes reversal in the vertical meridian, 
while it requires a + 1.50 D. to cause reversal in the horizontal 
meridian, what is the kind and amount of ametropia? 

In accordance with the theory of dynamic skiametry in 
which no allowance is to be made for the distance at which the 
test is made, this case would be one of compound hypermetropic 
astigmatism, showing 1 D. of defect in the vertical meridian 
and 1.50 D. of defect in the horizontal meridian, and the correct- 
ing lens would be + 1 D. sph. = + -50 D. cyl. axis 90°. 



An eye tested with the retinoscope by the customary static 
method shows the following: Vertical meridian point of reversal 
at 18 inches and horizontal meridian point of reversal of 32 inches. 
What is the refractive condition of each meridian, what two sphero- 
cylinders may be prescribed for distance, and which one would you 
consider best? 

Vertical meridian myopic 2.25 D., and horizontal meridian 
myopic 1.25 D. 

The correcting sphero-cylinders are: 

- 1.25 D. sph. = - 1 D. cyl. axis 190° 

- 2.25 D. sph. = + 1 D. cyl. axis 90° 

The latter would be considered best, because more periscopic 
and with a vertical axis. 



308 State Board Examinations 

With a + 1.50 in the trial frame, the 45th meridian of an eye 
shows the point of reversal at 27 inches, at the same time the motion 
in the 135th meridian is -with. By changing the + 1.50 to + 2.75 
the poi7it of reversal in this latter meridian is found to he at 40 inches. 
What is the refractive condition of each meridian, and what two 
sphero- cylinders may be prescribed? 

When the test is made at 27 inches, we must add — 1.50 D. 
to the lens that produces reversal, which in this case is equivalent 
to piano, showing the 45th meridian to be emmetropic. 

When the test is made at 40 inches, we must add — 1 D. to 
the lens that produces reversal, which in this case would show a 
hypermetropia of 1.75 D. in the 135th meridian. 

The correcting lens would be a piano-cylinder: 
+ 1.75 D. cyl. axis 45°, 
which, however, may be transposed into a sphero-cylinder: 
+ 1.75 D. sph. = - 1.75 D. cyl. axis 135' 



ro 



With the retinoscope at a distance of 40 inches from the eye 
as in the usual static method, a + 5.50 D. lens is required to neutral- 
ize the 120th meridian, and a — .50 D. to neutralize the 30th 
meridian. What is the amount of astigmatism present in the eye? 

This is ascertained by finding the difference between the 
two cylinders, or subtracting one from the other, which in this 
case, where one is plus and the other minus, would show 6 D. of 
astigmatism. 

Suppose the 15th meridian of an eye is emmetropic and the 
105th meridian is 2.50 D. hypermetropic; where is the point of 
reversal of each meridian when a — .50 D. lens is placed before the 
eye? 

In the emmetropic meridian the point of reversal is at 
infinity, and in the hypermetropic meridian beyond infinity. 
A — .50 lens would make the point of reversal in the emmetropic 
meridian beyond infinity, and in the hypermetropic meridian 
still further beyond. 

With a + 2.50 D. lens in the trial frame the point of reversal 
of the 30th meridian of an eye under test by the usual static method 



Retinoscopy 309 

of the retinoscope is found to he 32 inches. Changing the lens in 
the frame to a -\- 1.25 the point of reversal in the 120th meridian is 
found at 32 inches. What ki^zd of astigmatism is present and what 
is its aynotint? 

We must make allowance for the distance at which the test 
is made, adding a minus lens which corresponds to this distance. 
The lens which represents the working distance of 32 inches is 
1.25 D.; therefore, we must add — 1.25 to the finding in each 
meridian. Adding — 1.25 to + 2.50 would show the 30th 
meridian to be hypermetropic to the extent of 1.25 D. Adding 
— 1.25 to + 1.25 would show neutralization and indicate the 
120th meridian to be normal. 

Therefore it is a case of simple hypermetropic astigmatism, 
and the correcting lens is + 1-25 D. cyl. axis 120°. 



With a -\- 6 lens before the patient's eye the point of reversal 
for the 150th meridian is at 32 inches; for the 60th meridian at 16 
inches. What would he the correction for distance? 

To make allowance for the working distance of 32 inches, 
we must add — 1.25 D. to the neutralizing lens + 6 D., which 
shows a hypermetropia of 4.75 D. in the 150th meridian. For 
the distance of 16 inches w^e add — 2.50 D. to the same neutraliz- 
ing lens, which shows a hypermetropia of 3.50 in the 60th me- 
ridian. This would call for 

-f 4.75 D. cyl. axis 60° C + 3.50 D. cyl. axis 150°, 
or, 

+ 3.50 D. sph. C + 1.25 D. cyl. axis 60°. 



Working at a distance of 40 inches the retinoscopic finding for 
the 180th meridian of an eye is + 5 and for the 90th meridian it is 
+ 6. What would he the equivalent sphero- cylinders for distance 
correction? 

To make allowance for the working distance of 40 inches 
we must add — 1 D. to both meridians, which would show the 
180th meridian to be hypermetropic 4 D. and the 90th meridian 
hypermetropic 5 D., which would call for 



310 State Board Examinations 

+ 4 D. cyl. axis 90° C + 5 D. cyl. axis 180°, 
or, 

+ 4 D. sph. C + 1 D. cyl. axis 180°. 



Workiyig at a distance of 27 inches from the eye in the static 
method of retinoscopy, the 15th meridian needs a -\- 2.50 D. to get a 
choked disk, while the 105th meridian needs + 3.50 D. The full 
correction for distance is prescribed and torics are ordered. On 
testing the torics with a lens measure one surface is found to he a 
— 7.50 D. sphere; on the other surface the 15th meridian shows + 
8.50 and the 105th meridian shows -\- .50. Has the prescription 
been correctly filled? 

In working at 27 inches we must add — 1.50 D. to the lens 
that produces neutralization, which would show a hypermetropia 
of 1 D. in the 15th meridian and of 2 D. in the 105th meridian. 
In estimating the power of the toric lens we find that the concave 
surface against the convex curves, leaves just the proper amount 
of convexity in each meridian to correct its hypermetropia. 
Hence the prescription has been correctly filled. 



To what retinoscopic findings at one meter would a + ^ O — 
1.25 cylinder axis 90° correspond? 

This prescription w^hen analyzed shows a hypermetropia of 
2 D, in the vertical meridian and of + .75 D. in the horizontal. 
Inasmuch as working at one meter a — 1 D. must have been 
added to obtain this prescription, so now we must add + 1 D. 
to find the neutralizing lenses, which would show the retinoscopic 
findings to be + 3 D. vertically and + 1.75 D. horizontally. 



With a -\- 6 D. sphere before the patient's eye, the 135th. me- 
ridian shows the point of reversal at 40 inches, and the 45th meridian 
shows the same at a distance of 8 inches. What would be the distance 
correction? 

As the point of reversal in the 135th meridian is at 40 inches, 
we add — 1 D. to the + 6 D., showing a hypermetropia of 5 D. 
in this meridian. The point of reversal in the 45th meridian 



Retinoscopy 311 

being at 8 inches, we add — 5 D. to the + 6 D., showing a 
hypermetropia of 1 D. in this meridian. 

The correction for distance would be + 1 D. sph. O + 4 D. 
cyl. axis 45°. 

If the findings for distance by the retinoscope are + 1.25 for 
the 60th meridian and + 1.75 for the 150th meridian, what two 
sphero-cylinders may he prescribed? 

Assuming the examination was made at one meter, and 
adding —ID. for this distance, we find a hypermetropia of 
.25 D. in the 60th meridian and a hypermetropia of .75 D. in 
the 150th meridian. The correcting lens would be + -25 D. sph. 
.50 D. cyl. axis 60°, which could be transposed to + .75 D. sph. 
C- - .50 D. cyl. axis 150°. 



Change the retinoscopic findings -\- 2.50 D. in the 15th meridian 
at 40 inches, and + 3.50 D. in the 105th meridian at 32 inches to 
the proper sphero cylinder for distant vision? 

For the 15th meridian after making an allowance of 1 D. 
for the 40 inches, the refraction would be 1.50 D. hypermetropia; 
and for the 105th meridian after making an allowance of 1.25 D. 
for the 32 inches, the refraction would be 2.25 D. hypermetropia. 
The lens thus indicated would be + 1.50 D. cyl. axis 105° O 
H- 2.25 D. cyl. axis 15°, or + 1.50 D. S. C + .75 D. cyl. axis 15°. 



A toric lens used for distance measures on its convex surface + 8 
D. and on the concave surface it shows in the 30th meridian — 4.75 
D. and in the 120th meridian — 4.25 D. To what retinoscopic find- 
ings at 40 inches does this correspond? 




Fig. 34 



312 State Board Examinations 

A glance at the diagram shows that this toric lens has a power 
of + 3.25 D. in the 30th meridian and of + 3.75 D. in the 120th 
meridian. Assuming that the retinoscope had been used at the 
customary distance of one meter for which an allowance of 1 D. 
had been made, we must now add this same ID., which would 
make the retinoscopic findings + 4.25 D. in the 30th meridian 
and + 4.75 D. in the 120th meridian. 



With + 2D. the point of reversal for the 90th meridian is at 
40 inches, and for the 180th meridian 32 inches; what would he the 
full correction for distance? 

For the 90th meridian after making allowance of 1 D. for the 
40 inches distance, the correction is + 1 D., and for the 180th 
meridian after making allowance of 1.25 D. for the 32 inches dis- 
tance, the correction is + .75 D. 

This would call for + .75 D. cyl. axis 90° C + 1 D. cyl. 
axis 180° which is transposed to this sphere cylinder: 

+ .75 D. S. C + .25 D. cyl. axis 180°. 



A certain prescription is -{- 2 D. ^ -{-1.25 cyl. axis 180 . 
Working with the retinoscope at 32 inches, where will the point of 
reversal he for hoth meridians? 





+ 2. 




+ 3.25 




+ 1.25 


1+ 1.25 




+ 3.25 


+ 4.50 




+ 2. 


1 +2. 
1 + 1.25 








+ 3.25 


Sphero cyl. powers 


Retinoscopic findings 


Fig 


. 35 


Fig 


. 36 



The first diagram shows that the sphero cylinder has a power 
of + 2 D. horizontally and of + 3.25 D. vertically. And inasmuch 
as we must add 1.25 D. for the 32 inch working distance, the 
neutralizing lens would be + 3.25 D. for horizontal meridian and 
+ 4.50 D. for vertical meridian. 



Retinoscopy 



313 



The reversal point of a certain eye is for the 90th meridian at 
40 inches when a + 1.25 D. lens is used; for the 180th meridian 
testing at the same distance a + .75 D. is required. If a toric lens is 
prescribed, and on test with the lens measure one surface is found to 
he — 6 D. sphere, and the opposite surface to show + 6.25 D. in 
the 180th meridian and + 5.75 D. in the 90th meridian, has the 
prescription been properly filled? 

After making the usual addition of — 1 D., we find the 
refraction of the vertical meridian to be + .25 D., and of the 
horizontal meridian to be —.25 D., as shown in first diagram; 



+ .25 



- 6. 

4- 5.75 



- .25 



- .25 



Retinoscopic result 

Fig. 37 





- 6. 




+ 6.25 




+ .25 


ic lens powers 


Fig 


. 38 



An analysis of the toric lens shows — .25 D. power vertically 
and + .25 D. power horizontally as per the second diagram. 

A comparison of the two diagrams makes evident that the 
prescription has not been properly filled, but that the meridians 
have been reversed. 



With a-\- 1 D. lens before a certain eye the point of reversal for 
the 90th meridian is at 53 inches, and the point of reversal for the 
180th meridian is at 32 inches. What sphero cylinder will give full 
correction for distance? 

Adding — .75 D. for the working distance of 53 inches, the 
correcting lens for the 90th meridian is + .25 D. And adding 
— 1.25 D. for the working distance of 32 inches, the correcting 
lens for the 180th meridian is — .25 D. This would call for + .25 
D. cyl. axis 180° O — .25 D. cyl. axis 90°, which is transposable 
to the following sphero cylinder: -f- .25 D. sphere O — .50 D. 
cyl. axis 90°. 



314 State Board Examinations 

Can retinoscopy he practiced with the ophthalmoscope? 

Yes, it can, but the ophthalmoscopic mirror is concave, while 
a plane mirror is preferred. 



// a prescription calls for -\- 1 D. cyl. axis 90°, surface next to 
the eye to he deep concave, what kind of a lens would have to he made? 

A toric lens, which is built up from a base curve of 6 D. on the 
toric surface. The outside surface of such lens would show a + 6 
D. curve in the 90th meridian and a + 7 D. curve in the 180th 
meridian, while the deep concave surface towards the eye would 
be — 6 D. 

Is it true as claimed that retinoscopy is hest done with a plane 
mirror, and what is the reason? 

Retinoscopy can be done with either the plane or concave 
mirror, but the plane is preferred, because in its use the distance 
at which the test is made need not be considered, where with the 
concave mirror care must be taken to keep beyond the focal dis- 
tance of the mirror. 



With a plane mirror at 20 inches in the usual static method of 
retinoscopy, a -\- 1 D. sphere neutralizes motion .in all meridians; 
what is the full correction for distance? 

In working at the customary distance of one meter, an allow- 
ance of 1 D. is made: so in working at 20 inches an allowance of 
2 D. must be made. 

We, therefore, add — 2 D. to + 1 D- and the results is 
— ID., which would be the full correction for distance. 



In a certain person the error of refraction is corrected by the 
following formula: + 0.50 D. S. = -{-ID. cyl axis 45°. In the use 
of the retinoscope with the usual static method, what will he the move- 
ments of the fundus reflex at a distance of one meter without lenses? 

This formula would indicate that the condition of the refrac- 
tion was hypermetropic, one meridian to a greater extent than 
the other, and that therefore the emerging rays would be diver- 



Retinoscopy 315 

gent and would not meet in a focus ; as a result of which there would 
be no neutral point between the eye under examination and the 
observer. Hence, the movement of the reflex in the pupil with the 
plane retinoscope at a distance of forty inches and without any 
lenses before the patient's eye, would be with. If a concave mirror 
was used the movement would be against. 



What are the optical principles of retinoscopy with a plane 
mirror? 

In retinoscopy the observer directs his attention to the move- 
ments of the shadowy edge of an aerial image of a bright spot 
formed from the light thrown by a mirror on the fundus of the 
observed eye, as it appears to pass across its pupil. 

The direction of the movement in the pupil of the observed 
eye as it appears to the observing eye, will depend upon the posi- 
tion of the latter relative to the image formed at the conjugate 
focus of the fundus of the observed eye. Or in other words 
whether the observer receives rays which come from an aerial 
image of the bright spot, or the rays which proceed from the bright 
spot itself, before they have been united to form an image. 

In this latter condition where the conjugate focus would fall 
behind the nodal point of the observing eye, the movements are 
always in the same direction as the mirror rotation, just as the 
retina of the eye would perceive the motion of any object in front 
of it. 

But when the emerging rays from the observed eye are 
brought to a focus in front of the observing eye, the movement in 
the pupil will always be in a direction opposite to that of the mirror 
rotation. 

If the observer places his eye in such a position that the conju- 
gate focus of the observed eye falls upon it, no movement will be 
apparent. 

In determining the refraction of an eye by retinoscopy, we 
find that the far point of the eye or the conjugate focus of its 
fundus, which we can do by artificially bringing the far point to 
any derived finite distance by means of a lens placed in front of 
the eye. 

In myopia of 1 D. the fixed far point is at one meter, which is 
the position of the observing eye. The far point of any eye can 



316 State Board Examinations 

be brought to the same point of reversal by the proper lens placed 
in front of it. 

The difference between the actual far point and 1 D. of my- 
opia, is represented by the lens that was found necessary to bring 
the rays to a far point at one meter. Therefore, the far point is 
expressed by the difference between the number of the lens 
employed and ID. 

In other word we add — 1 D. to the neutralizing lens. For 
example if + 1 D. was found necessary for neutralization, then 
+ 1 D. — 1 D. = or emmetropia. 

If neutralizing lens was + .50 D. then + .50 D. — 1 D. = 
.50 D. or half diopter of myopia. 

If the neutralizing lens was — 2D. then — 2 D. — 1 D. = 
— 3 D. or three diopters of myopia. 

The accommodation of the observed eye should be passive, 
as otherwise it is impossible to measure its static refraction. 

But accommodation on the part of the observing eye need 
not be taken account of, as its use or non-use cannot change the 
result. 

A patient has 2 D. of myopic astigmatism. If a + 2D. sphere 
he placed before the eye, where will the point of reversal he for the 
horizontal meridian if the cylinder required to correct the error must 
he placed axis vertical? 

If the error of refraction is simple myopic astigmatism, w^hich 
is corrected by a — 2 D. cyl. axis 90°, we know the vertical 
meridian of the patient's eye is emmetropic, and the horizontal 
meridian myopic to the extent of 2 D. 

When a + 2 D. sphere is placed before such an eye, the emme- 
tropic vertical meridian will be made artifically myopic to the 
extent of 2 D. and the point of reversal w411 be 20 inches. 

At the same time the natural myopia of the horizontal 
meridian will be artificially increased to 4 D. in which case the 
point of reversal will be at 10 inches. 



Using the retinoscope at 40 inches with a -\- 1 D. sphere hefore 
the eye hei^ig examined, the horizontal meridian shows neutrality. 
When the + ID. sphere is removed, the vertical meridian shows 
neutrality. What is the error of refraction? 



Retinoscopy 317 

In working with the retinoscope at 40 inches, if a + 1 D. lens 
neutralizes the movement, emmetropia is indicated; therefore, in 
this case we must assume that the horizontal meridian is emme- 
tropic. 

When neutrality of movement occurs naturally without the 
intervention of any lens, myopia is indicated, the amount of 
which would correspond to the distance of the neutral point. In 
working at 40 inches, neutrality of movement would indicate 
m^^opia of 1 D., and therefore, in this case the vertical meridian 
is myopic ID. 

If, then, the horizontal meridian is emmetropic, and the ver- 
tical meridian myopic ID., the error of refraction is simple myopic 
astigmatism, and the correcting lens would be — 1 D. cyl. axis 
180°, which corresponds to astigmatism with the rule. 



Physiology of Vision 

What prevents an excessive amount of light from entering the 
eye? 

It is the function of the iris to regulate the amount of Hght 
entering the eye, owing to the fact that exposure to hght causes 
the circular fibers of the iris to contract and as the light increases 
in intensity the pupil assumes its smallest possible size, this con- 
traction being the greatest when the light falls upon the macula. 



At what age is the pupil of the eye the largest? 

The size of the pupil is influenced by age, the color of the iris 
and the character of the refraction. Other things being equal the 
pupil is apt to be larger in youth, in dark eyes and in myopic eyes. 
The average diameter of the pupil is about 4 mm. and while we 
expect them to be of the same size in the same individual, yet 
slight differences in the width of pupils are not incompatible with 
health. 

State the character of the retina. Give the relation of the retina 
to the optic nerve. 

The retina is a continuation or expansion of the optic nerve. 
It is the nervous and percipient coat of the eye, and is intended to 
receive the images of external objects and transmit them to the 
brain. It is most sensitive in the region of the macula, and least 
sensitive near the ora serrata. It is transparent and hence invisi- 
ble in health. 

Describe a normal fundus . 

The fundus is reddish in color, due to the choroid coat. 

The optic disk is the most prominent feature of the fundus. 
It is about 1.5 mm. in diameter and is situated inwards from the 
posterior pole of the eye. It may be round, but is usually slightly 
oval vertically. 

318 



Physiology of Vision 319 

The central artery of the retina and its vein pass through the 
axis of the optic nerve. Near the center of the disk is usually seen 
a depression known as the physiologic cup. 

The scleral and choroidal rings may be seen around the disk 
of light and dark color respectively. 

The color of the disk is pinkish and distinctly lighter than 
the surrounding fundus. The veins are recognized by being larger 
in size and darker in color than the arteries. 

The macula lutea is situated on the temporal side of the disk. 
It is oval in shape with its long diameter horizontal and it is 
darker in color than the rest of the fundus. At its center is the 
fovea centralis, which gives a reflex and appears as a bright spot. 



> Differentiate between the range and the power of accommodation. 

The range of accommodation is the distance between the 
near point and the far point while the power of accommodation is 
the force which the ciliary muscle is able to exert to adapt 
vision for the near point. 

What is accommodation and how is it produced? 

Accommodation is the change in the dioptric condition of the 
eye whereby it is adapted for the perception of objects close at 
hand. The formation of a perfect image upon the retina depends 
upon the dioptric apparatus of the eye being so adjusted that the 
rays, diverging from any particular point, shall be brought to a 
focus and form an image upon the retina. For the parallel rays 
proceeding from objects at a distance, the static refraction of the 
eye suffices. From objects closer than twenty feet, the rays are 
divergent, the amount of the divergence increasing with the near- 
ness of the object, and now the static refraction is insufficient to 
form a clear image, and the dynamic refraction is brought into 
play by means of the accommodation, and for each different dis- 
tance there must be a change in the accommodation, increasing 
as the object approaches and diminishing as the object recedes. 



What is amplitude of accommodation? 

This is the power of accommodation which the eye is able 
to exert and represents the difference in the dioptric power of 



320 State Board Examinations 

the eye when the accommodation is passive and when at its 
maximum activity, or the difference between the static and the 
dynamic refraction of the eye. The ampHtude of accommodation 
is greatest in youth and gradually diminishes with age, until at 
seventy- years of age it is entirely lost. About middle life this 
amplitude is so impaired that near vision is \-ery much interfered 
with; this condition of presbyopia calls for the assistance of a 
convex lens to supply the deficiency. 



What is converge7ice? By what means is it obtained^ 

Convergence is the act of directing the visual axes of the two 
eyes to the same point at some near distance, in order that the 
images formed in the eyes shall be caused to fall on the macula 
of each and be fused so that one object is seen, instead of two. 
Double vision results when the image falls upon parts of the two 
retinae which do not exactly correspond. The nearer an object 
approaches the eyes, the more strongly they must be converged, 
while at great distances the need of convergence diminishes cor- 
respondingly. 

The function of convergence is accomplished by means of 
the internal recti muscles, which are also known as the muscles 
of convergence. 

What constitutes the amplitude of convergence? 

The amplitude of convergence is the power of convergence, 
or the whole amount that can be produced by the strongest effort 
of the internal recti muscles, and it is usually expressed in meter 
angles. At extreme distances the visual lines are assumed to be 
parallel ; when the eyes are directed to an object one meter away, 
the visual lines must converge to this point, thus forming the 
meter angle, which in this case, equals one and may be expressed 
as follows : C = 1 (the C being the sign for amplitude of conver- 
gence.) 

If the object looked at be situated at half a meter, the angle 
will be twice as great, and then C = 2. If situated at one-third 
of a meter, the angle of convergence is correspondingly increased, 
and then C = 3. If the object is removed two meters or four 
meters, the angle is diminished in proportion, and then C = 1 2 
or C = 1,4. 



Physiology of Vision 321 

What is the visual field? The visual line? 

The visual field is the circular space in front of each eye, 
within which objects are distinctly perceptible while the eye is in 
a fixed position. In some of the lower animals, on account of the 
position of the eyes, the field of vision is a complete sphere, objects 
being visible in every direction. In man the field is limited, and 
yet it amounts to almost 180° because if the object is brilliant and 
placed laterally at the external borders of the field, it does not 
escape perception. 

Within this field, however, there is only one point where 
objects can be seen with perfect distinctness, and that is in the 
center of the field, and its prolongation forward from the pupil 
is known as the "line of direct vision." 

The limitation of distinct sight to the line of direct vision is 
practically compensated for by the great mobility of the eyeball, 
which rapidly turns in all directions, shifting the line of vision and 
examining, in turn, every part of the field of vision. 



What is meant hy binocular vision? 

When the two eyes are directed to an object, an image is 
formed in each and the function of convergence which presides 
over this department of vision so directs the visual axes of the 
two eyes that the image is formed on the macula of each eye or, 
at least, on "corresponding portions" of each retina. jThe impres- 
sion is carried to the brain by each optic nerve, but as the images 
are exactly alike and are formed on parts of the tw^o retinae that 
correspond, they are so combined by the brain as to give the 
impression of a single object. This simultaneous use of both eyes, 
resulting in the production of single vision, is known as binocular 
vision, which may be briefly defined as single vision with two eyes. 



What is the function of the choroid? 

The choroid is the vascular coat of the eye and consists of 
connective tissue stroma which supports the numerous blood- 
vessels, the larger of which are in the outer layer, while the inner 
layer is devoted to the capillary netw^ork. Therefore, the choroid 
is to be considered the chief nutritive apparatus of the eye, be- 



322 State Board Examinations 

cause the percipient layers of the retina and the ciUary muscle, 
which are the most active portions of the organ of vision, receive 
their nutrition from this source. It also contains pigment matter, 
which protects the eye from excessive light. 



In what way are sight impressions received and conveyed to 
the brain? 

A light wave starts on its journey from the luminous object 
from which it originates. After undergoing the proper refraction, 
the first actual step towards becoming a visual impulse is taken 
when it decomposes the photo-chemical substances of the retina 
and sets up vibrations in the extreme periphery of the end organs 
of this membrane. 

The excitation received by the retina is then conveyed by 
fibers of the optic nerve back to centers at the base of the brain 
and thence to the visual cortex, which is that part of the con- 
voluted cerebral surface concerned with the function of sight. The 
result in the cortical centers receiving the impulse is a visual 
sensation or perception. 



Why is it necessary for the aqueous humor to he thinner than 
the vitreous, and yet it has the same density? 

These two humors have practically the same index of refrac- 
tion, viz., 1.33, but they do not have the same consistency, the 
aqueous being quite thin and w^atery, while the vitreous is semi- 
fluid, gelatinous or jelly-like. One reason w^hy the aqueous should 
be thinner is that the movements of the iris may quickly take 
place, with nothing to retard them, as would be the case if the 
humor in which they floated was thicker. 



Why are accommodation and convergence closely related? - 

The purpose of the latter is to form a clear image upon the 
retina, while the former directs the two eyes to the same point, so 
that the image may fall upon the macula of each eye, in order that 
single binocular vision may result. For every diopter of accom- 
modation there is a corresponding meter angle of convergence for 



Physiology of Vision 323 

each eye. In order that this relation may be maintained, nature 
has very properly provided that the ciliary muscle and the inter- 
nal recti muscles shall be supplied by the same nerve, the third 
cranial or oculo-motor nerve. As the years pass the habits formed 
by the accommodation and convergence become more and more 
fixed and the relationship made still closer. This association of 
these two functions is not the same in all people, and as the pres- 
byopic period gradually approaches new relations must be formed. 



What influence has the age upon the accommodation? 

The accommodation depends principally upon two factors — 
the contractility of the ciliary muscle and the elasticity of the 
crystalline lens. i\s age advances the ciliary muscle gradually 
loses its power of contractility and the lens its power of elas- 
ticity. These two changes, the loss of strength of the muscle, 
and particularly the increasing hardness of the lens, have neces- 
sarily a restricting influence upon the accommodation. The in- 
crease in the density of the lens makes it more dif^cult to be acted 
upon by the muscle for a change of curvature, even though the 
latter retained all of its primary strength. 

The effect of age upon the accommodation is modified by the 
condition of the refraction of the eye ; in hypermetropia the lessen- 
ing of accommodative power is intensified and made manifest 
earlier, whereas in myopia the accommodation is less impaired 
and, then, not so early in life. 



What is the function of the eye-lashes? 

When the lids are partly closed the lashes come together in 
such a way as to form a kind of screen, which, w^hile not excluding 
vision, serves as a protection against wind and dust and light and 
all foreign bodies floating in the air. 



In what way do they differ from ordinary hairy growths? 

Their bulbs are freely supplied with nerves, giving them tac- 
tile sensibility, so that they are enabled to act as "feelers" to 
warn the eye and reflexly cause the lids to close tightly on the 



324 State Board Examinations 

approach of any small object, as an insect in the dark or when the 
vision is not on guard. 



What arrangement is there on the posterior surface of the iris 
to prevent the transmission of light? 

A pigment layer covering the posterior surface of the iris as 
far as the anterior margin of the pupil. 



In what way do the nerve filaments end in the retina? 
In the layer of rods and cones. 



Which of these is most highly developed in man? 
The rods, which are also by far the most numerous. 



Which are most numerous of the macula lutea ? 

The cones ; in the fovea centralis they alone are present. 



What are entoptic images? 

The word "entoptic" is derived from the Greek and means 
within the eye. These images depend upon the presence of some 
opacity in the transparent media of the eye, such as muscse 
volitantes. They are to be found in all eyes to a certain extent, 
and are made more evident when looking at a white cloud or any 
light-colored object, especially through a pin hole. 



What is meant by intraocular pressure? 

The tension to which the coats of the eye are put by the 
varying quantity of the humors of the eye. When the tension is 
abnormally increased, the condition is known as glaucoma, a 
disease that is often fatal to vision. 



What is the visual purple? 



Physiology of Vision 325 

A certain coloring substance of the retina found in the exter- 
nal segments of the rods, the color being uniformly distributed 
throughout this portion of the rod, while it is absent from the 
cones and the macula lutea, and hence it cannot be considered 
essential to the act of vision. But it undergoes changes when 
exposed to light and, therefore, it is probable that it plays some 
important role in the visual process. The visual purple disappears 
when the eye is exposed to light and is restored when the light is 
excluded. It can be seen by the naked eye or the microscope under 
the light of a sodium flame in the fresh retinae of animals which 
have been kept for an hour or two in the dark. When light acts 
upon the visual purple, it first produces visual yellow and then 
visual white which latter is described by scientists as a greenish, 
fluorescent substance. 



What relation does the convergence of the eyes hear to the accom- 
m odation ? Expla in fully . 

In the normal eye the relation between convergence and ac- 
commodation is very close. For every effort of one there is a cor- 
responding effort of the other. In emmetropia at one meter there 
is required 1 D. of accommodation and 1 meter angle of conver- 
gence for each eye; at one-half meter there is required 2 D. of 
accommodation and 2 meter angles of convergence for each eye 
and so on. 

Although so closely associated, the relation between accom- 
modation and convergence is not entirely rigid; within certain 
limits either may be varied slightly. The accommodation may be 
lessened by convex lenses or increased by concave lenses, without 
any change in the convergence. This relative range of accommo- 
dation varies in dift'erent individuals, from 3 D. in distant vision 
of a young emmetrope to 6 D. in near vision. 

As the accommodation can be thus varied without change of 
the convergence so the latter within certain limits can be increased 
or diminished without affecting the accommodation. The conver- 
gence can be diminished by a prism, base in, when the accom- 
modation is completely relaxed. It can also be increased by 
means of prisms, bases out, without bringing the accommoda- 
tion into action. This relative range of accommodation varies 
in different persons up to 6 m. a. 



326 State Board Examinations 

The flexibility in the relation between accommodation and 
convergence is of importance as allowing comfortable binocular 
vision in ametropia that would otherwise be impossible. 



What is the effect of the contraction of the radiating muscles of 
the iris? 

Dilate the pupil. 




Candle Flame Images in Eye 

Fig. 39 

What effect would he produced hy cutting (a) the superior oblique 
muscle, (b) the internal rectus muscle? 

(a) The superior oblique muscle turns the eye downward 
and outward, rotating the upper end of the vertical meridian 
inward, hence cutting of this muscle would limit the movement 
downward and outward with a corresponding strabismus upward 
and inward. 

(b) Cutting of the internal rectus muscle would cause limi- 
tations of the power of adduction, with perhaps, divergent 
strabismus and crossed diplopia. 



Physiology of Vision 327 

How is the tension of the eyeball regulated? 

The tension of the eyeball is maintained by the secretion of 
the aqueous humor, and its equilibrium is regulated by the escape 
of this humor into the spaces of Fontana, the canal of Schlemm 
and the anterior ciliary veins. 



What effect would he produced if the sixth nerve were severed? 

The sixth cranial nerve is called the abducens and supplies 
the external rectus muscle of the eyeball. Severence of this nerve 
would cause inability to turn the ball outwards, with perhaps con- 
vergent strabismus and homonymous diplopia, which would be 
increased by looking toward this side and diminished when look- 
ing toward the opposite side. 



Why is the optic disk a blind spot in the eye? 

It is insensitive to light because the percipient elements of 
the retina are lacking here. 



What is emmetropia and in what way does it differ from ame- 
tropia? 

Emmetropla signifies an eye in measure, so that parallel rays 
are exactly focused upon the retina without any effort of accom- 
modation. 

This is in contrast with hypermetropia, where the focus of 
parallel rays is behind the retina and with myopia, where it is 
in front. In simple astigmatism the focus of one meridian is on 
the retina and of the other back or in front of it. In compound 
astigmatism the focus of both meridians is in front or back of the 
retina, at varying degrees. 



What is the range of focus between an eye at rest and one under 
full accommodative power called? 

The range of accommodation. 



328 State Board Examinations 

What is meant hy the term visual acuity? 

Sharpness of sight or the amount of vision possessed as com- 
pared with a standard, and has reference to the abiHty to perceive 
and recognize form and outUne. 



What is anisometropia? 

That condition in which the refraction of the tw^o eyes varies. 



In high degrees of anisometropia what is the effect of hinocidar 
vision? 

If the refractive difference is not too great, or if the patient 
can equalize the refraction of the two eyes by an unequal con- 
traction of the ciliary muscles then binocular vision is likely to 
be present. But if the difference is considerable and the ciliary 
muscles receive the same amount of innervation so that the rela- 
tive difference between the two eyes is constantly maintained the 
retinal image in the more defective eye is not only dimmer, but 
of a different size from its fellow under which circumstances the 
incentive to binocular vision is very much weakened and finally 
lost, or, if binocular vision is maintained, it is at the expense of 
constant effort. In other cases of greater inequality of the retinal 
images, the fusion sense is so weakened that no effort is made to 
maintain binocular vision. The imperfect image of the poorer eye 
is ignored or suppressed by the brain, and then monocular vision 
is said to exist. 

What is the yellow spot of the eye? 

It is a highly sensitive area of the retina, situated at the back 
of the eye, in the Hne of the visual axis about 1.5 mm. to the tem- 
poral side of the posterior pole. This area is about 2.5 mm. in 
diameter, and is also known as the macula lutea, or simply the 
macula. But the true macula is only about 1 mm. in diameter, and 
is the area of most distinct vision. In the center of the macula 
is a minute depression about .25 mm. in diameter, consisting 
entirely of narrow cones packed closely together, w^hich is called 
the fovea centralis on account of its position, and it is here that 
vision is most acute. 



Physiology of Vision 329 

What is the cause of convergence? 

The abhorrence of diplopia or the desire implanted by Nature 
for single vision and, as a result, the innervation of the internal 
recti muscles, which are the muscles concerned in the function of 
convergence. 

What is the blind spot of the eye? 

It is located at the entrance of the optic nerve, where the 
fibers pass through the sclerotic at the back of the ball, about 
2 mm. to the nasal side of the posterior pole. This optic nerve 
head, on account of the absence there of the retina proper, is 
totally insensitive to light, and hence it is known as the blind spot. 

We are not conscious of the existence of this blind spot be- 
cause when our eyes are directed toward an object the image is 
formed upon the macula, which is in the line of direct vision, 
while the blind spot is situated to the inner side of this point. 

When both eyes are open an object may be so placed that 
its image falls upon the blind spot of one eye, but in such case it 
must fall upon the macula of the other eye, and hence the object 
will be distinctly seen. 

It is impossible that an image should fall upon the blind spot 
of both eyes at the same time. 

Even when one eye only is used this blind spot is not notice- 
able because it is located in a part of the field to w^hich our attention 
is seldom directed, and where the perception of objects is so 
imperfect that the absence of one of them momentarily is not 
regarded. 

What is the monocular field of vision? 

W^ith the eye looking straight forward (and confined to one 
eye) it is the space within which objects are visible. Vision is per- 
fect only at the center in the line of direct vision, and becomes less 
distinct towards the periphery of the field. 



Why is binocular field of vision wider than the monocular kind? 

Because the second eye gives all the additional field on its 
side, which is not possible to one eye on account of the projection 
of the nose. 



330 State Board Examinations 

\l 
What is the principal advantage of indirect vision? ^ 

Deprived of indirect vision would place a man in the position 
of looking through a long, narrow tube, which would allow of 
seeing nothing but the object in the line of direct vision. It 
would, be impossible to see objects on one side or the other, 
above or below, without an incessant turning of the head. As 
a person walks across the street looking straight ahead to the 
opposite side, where he is going, he is able without turning his 
head or eyes, to see if any vehicles are approaching in either 
direction or if there are any obstructions or depressions in the 
street by means of his indirect vision, the importance of which 
can hardly be overestimated, as it enables the man to avoid 
dangers which approach and menace him from all sides. 



What causes the crystalline lens of the eye to become more 
convex when looking at a near object? 

The increase in convexity of the crystalline lens is accom- 
plished by means of the contraction of the ciliary muscle. 



What is diplopia and how does it differ from heterophoria? 

Heterophoria is an imbalance of the ocular muscles ; diplopia 
is double vision. The first is a condition, the second, a symptom. 
We may have and usually do have heterophoria without diplopia 
because the eyes are able to overcome the imbalance, but w^e 
do not have diplopia without some disturbance of the muscular 
equilibrium. 

Heterophoria must be sought and discovered by the tests of 
the optometrist, while diplopia is self-evident to the patient 
himself. Heterophoria may be described as a latent condition 
while diplopia is always manifest. 



What is the difference between binocular vision and fusion? 

Binocular vision is single vision with two eyes, and depends 
upon the blending in the brain of the impressions that are made 
upon corresponding parts of the two retinae. This usually 
exists in connection with fusion, as shown in the use of the 



Physiology of Vision 331 

stereoscope, where if the pictures are separated or approximated, 
the eyes will follow them in the interest of binocular vision in 
the effort to maintain fusion. 

It is conceivable that there may be some cases of binocular 
vision of such grade that the two pictures of the stereoscope will 
be united in one only when placed in certain relative positions 
corresponding to the directions independently assumed by the 
visual axes, showing no effort to produce or maintain fusion on 
account of an absence of desire for binocular vision. 



What is binocular vision? / 

Single vision with two eyes, for the maintenance of which 
it is necessary that the images of the object fall upon identical 
portions of the two retinae in order that they may transmit to 
the brain a single impression; or the object must be at the point 
of intersection of the two lines of direct vision. 

Vision of one eye gives a flat appearance, while binocular 
vision gives the impression of depth and solidity, as well as a 
more correct estimate of distance on account of the amount of 
convergence that is brought into play. 



Describe the mechanism of accommodatio7i. 

This depends upon the elasticity of the crystalline lens and 
the contractility of the ciliary muscle. The latter has such 
connections with the former that when the muscle contracts 
the lens becomes more convex, this increase of convexity being 
greater on its anterior surface, and when the muscle relaxes the 
lens becomes less convex. There are two theories of accom- 
modation— Helmholtz's and Tscherning's — but the essential 
feature is that accommodation depends upon an increase in the 
convexity of the crystalline lens and is accomplished by the 
contraction of the ciliarv muscle. 



What is the difference in meaning between amplitude of accom- 
modation and range of accommodation? 

The range of accommodation is the distance between the 
far point and the near point, and is the distance over which the 



332 State Board Examinations 

eye has command by aid of its accommodation. The ampHtude 
of accommodation is the power of accommodation or the force 
necessary to change the adaptation of the eye from its far point 
to its near point, and it is represented by the difference in the 
refractive power of the eye when in a state of complete rest and 
when at its maximum of accommodation. 



What is the function of the cornea? 

The function of the cornea is two-fold : 

It is part of the external coat, and as it is tough and unyield- 
ing, it helps to maintain the shape of the eyeball and protect its 
contents. 

It is also one of the refracting media of the eye, receiving 
the rays of light, and by its density and convexity converges 
them towards the next medium. 



What is the function of the iris? 

To regulate the amount of light admitted to the retina. In 
a bright light the sphincter fibers contract to protect the eye from 
the excess of light; in a darkened room the dilator fibers contract 
to allow the entrance of as much as possible of the insufficient 
light. 

What is the function of the crystalline lens? 

To still further converge the rays of light entering the eye, 
and by its varying convexity to focus them upon the retina no 
matter what the distance of the object may be from which they 
come. - — 



What is the result of the functioning of the ciliary muscle? 

An increase in the convexity of the crystalline lens as in the 
act of accommodation, so as to adapt the eye for the diverging 
rays proceeding from close objects. 



What is the purpose of the extrinsic muscles of the eye? 



Physiology of Vision 333 

To turn the eyeballs in the various directions that may be 
necessary and for the maintenance of binocular vision. 



What is the difference between supraduction and infraduction? 

Supraduction is a turning upward; infraduction a turning 
downward. 

The crystalline lens is sometimes called a humor; is this correct? 

The eyeball is described as being composed of three coats 
within which are contained three humors. These humors are 
the aqueous, the crystalline and the vitreous. The crystalline is 
denser and thicker than the other two humors, but at the same 
time it is jelly-Hke or semi-fluid, so that it is really a humor. 



What is the usual difference between the pupillary distance for 
far and that for near? 

The pupillary distance for reading is about four mm. (one- 
sixth of an inch) less than for distance. 



How can you find the number of degrees of convergence used at 
the ordinary reading distance? 

The degree of convergence is expressed in terms of the meter 
angle, which is the angle through which each eye must turn from 
parallelism of the visual lines so that these lines may meet at a 
distance of one meter. The advantage of this system is that in 
emmetropia the meter angles of convergence are equal to the 
diopters of accommodation. The objection to it is that the meter 
angle has no fixed value on account of the variation in the inter- 
pupillary distance, but for the average distance it is about 1>2°. 
As the angle of deviation is about one-half the refracting angle of 
a prism, this would correspond to the deviation of a 3° prism. 
Therefore, one meter angle of convergence is equivalent to the 
effect of a prism of 3° before each eye, or to the effect of a prism of 
6° before one eye. 

At the usual reading distance of 13 inches, where 3 D. of 
accommodation is in use, the equivalent would be 9° before each 
eye or 18° before one eye. 



334 State Board Examinations 

\J 
Which is the stronger function, accommodatioji or convergence? 
On what do you base your conclusion? 

There may be a difference of opinion on this point, and it 
probably varies in different persons and at different times of life. 
There is always an abhorrence of diplopia, which is more dis- 
turbing than a sHght indistinctness of vision, and would seem to 
indicate that convergence was the most important. But, on the 
other hand, in young hypermetropes in order to bring the accom- 
modation into play and secure clear vision strabismus and mo- 
nocular vision are established, which would seem to indicate that 
accommodation was the dominant function. 



What are the two associated functions of the eye as regards 
changes of the gaze from point to point, and what is the nature of 
this association? 

Accommodation and convergence, the association between 
which being so close that exercise of one of them is involuntarily 
accompanied by a corresponding action of the other. 



What stimulus is it that results in exact binocular vision? 

The fusion sense or fusion faculty, aided by a strong natural 
desire for single vision. 



Name all of the extrinsic muscles brought into play in converging 
to a point eighty inches away; four inches away. 

Converging at eighty inches is accomplished mainly by the 
internal recti muscles. At four inches these muscles are reinforced 
by the superior and inferior recti muscles. 



When is the punctum proximum of accommodation and con- 
vergence at the same point? 

In emmetropic and orthophoric eyes, when the accommoda- 
tion and convergence can be used in the same proportion and to 
the full extent to focus and fix a near object. 



Physiology of Vision 335 

In what way does the vision of a color-blind person differ from 
the vision of a person with normal vision? 

Color blindness does not seem to reduce the visual acuity as 
it might be thought to do at first sight, but as a matter of fact the 
general vision of the color blind is up to the standard of normal 
eyes. In those colors which they cannot see as such they can dis- 
tinguish differences of shade and tone that are dependent upon 
the admixture of white, even better than normal eyes. 

When a totally color blind person looks at a colored object he 
is conscious only of the white light which is present in varying de- 
grees in every color, and instead of color the object presents shades 
of gray, as in an engraving. In cases of blindness for red, when a 
red object is looked at there is no stimulation of the red fibers, 
but there is an impression of the green fibers, and to a slight extent 
of the violet. A red object therefore makes the same impression 
as^ green one, but he is sometimes able to distinguish between the 
two by their difference in brilliancy. 



Why are we not conscious of the existence of the blind spot when 
we close one eye? 

Because it covers such a small area, and besides is located in 
a part of the field of vision to which our attention is scarcely 
directed, and where the perception of various objects is so imper- 
fect that the momentary absence of one of them is not regarded ; 
and finally because the brain from long experience has learned to 
ignore it. 

What is the purpose of the iris of the eye and to what extent is 
it effective? 

To regulate the amount of light admitted to the eye. There 
is a limit to its contractibility in shutting out excessive light, as 
there is to its dilatability in admitting diminished light. 



What is the difference in the dioptric power of the eye between 
when the accommodation is at rest and in full force called? 



336 State Board Examinations 

It is that which is suppHed by the amplitude of accommoda- 
tion, which, added to the power of refraction of the eye when in 
repose, represents its full positive refracting power. 



What effect does age have on the accommodation and on the con- 



vergence? 



There is a steady and gradual diminution of the power of 
accommodation with the advance of years, while the power of 
convergence is but little if any affected. 



What is meant hy positive and negative convergence, and what is 
the total amplitude of convergence? 

Positive convergence is the turning inward of the visual lines 
so that they shall meet at the point of fixation. 

Negative convergence is a divergence of the visual lines. 

The total amplitude of convergence is represented by the 
nearest point for which the eyes can converge. If this point was 
5" inches, it would represent a converging power of about 8 meter- 
angles. 

What is meant hy the term binocular vision? 
Single vision with two eyes. 



In what way does the human eye resemble a photographic camera 
and in what way is it different? 

In the human eye the image is formed on the retina by means 
of the crystalline lens which is adjustable for vision at different 
distances, very much as in the camera, the image is formed on the 
sensitive plate or film by means of the converging lens, the instru- 
ment being adjustable for objects at different distances, by alter- 
ing the distance between the lens and the film, which is capable 
of receiving only one image, while the capacity of the retina is 
unlimited. 

Why does the pupil of the eye look black when it is really 
transparent? 



Physiology of Vision 337 

Because there is no intraocular illumination, or no inside 
light to come out. If light entered the eye from a flame, it 
would return in the same direction from which it came, and in 
order for the observer to see it, he must get in the path of the 
returning rays, and in so doing his head gets in the way and shuts 
off the entering light. This is overcome in the ophthalmoscope 
where the mirror acts as the source of light, and in returning 
some of the light passes through the sight hole into the eye of the 
observer. 



What is the difference between monocular vision, binocular 
vision and the fusion sense?^ 

Monocular vision is one eye vision. In binocular vision 
both eyes are used and the image formed on the macula of each, 
but they are blended in the brain so that we are conscious of 
only#one object. 

The fusion sense may be regarded as presiding over binocular 
vision, and is that function of the visual centers in the brain 
that enables fusion of the two images to be m.ade and maintained. 



How is the accommodation of the eye produced and why does 
it fall off with age? '--■ 

The accommodation is produced by an increased convexity 
of the crystalline lens and is caused by contraction of the ciliary 
muscle. It lessens with age because the crystalline grows denser 
and harder and is not able to respond so readily to the action of 
the ciliary muscle. 

What is the range of accommodation in any individual, and 
how may it be measured? 

The writer regards the distance between the near and the 
far points over which the eye has command, by the aid of its 
accommodation, as the range of accommodation. 

But the question implies what the writer terms the ampli- 
tude of accommodation, which is the force necessary to change 
the eye in its adaptation from its far point to its near point and 
is represented by that convex lens which would enable the eye 



338 State Board Examinations 

to see at its near point when the accommodation is at rest. The 
accommodation is therefore equal to a convex lens if such strength 
as will give to rays proceeding from its near point the same 
direction as if they come from the far point. Hence we measure 
the near point and transpose into D's. As for instance a near 
point if ten inches represents an accommodation of 4 D. 



What is meant by the term spasm or cramp of the accommoda- 
tion? 

A persistent contraction of the ciliary muscle, which fails 
to relax, even when there is no need for its contraction. 



What is the difference between subnormal accommodation and 
subnormal range of accommodation? 

By subnormal accommodation the writer would understand 
that the power or amplitude of accommodation was below the 
normal standard for that particular age. 

By subnormal range of accommodation we would understand 
that the distance between near and far points is less than normal. 



How close is the relation between accommodation and conver- 
gence? 

This relation is close but not absolute, because although so 
intimately connected, they may within certain limits be used 
independently of each other. 

For instance, the effort of accommodation may be increased 
or diminished by the use of concave or convex spheres, respec- 
tively, while the same degree of convergence is maintained. 

Or the effort of convergence is lessened by prisms bases in 
without a corresponding diminution of accommodation, as shown 
by the fact that the distinctness of the object is not impaired. 

Or the effort of convergence may be increased by prisms 
bases out without a corresponding increase of accommodation, 
as shown by no interference with the clearness of vision. 

But at the same time the general statement holds good that 
the functions are so closely related that with every effort of 
accommodation there is a corresponding effort of convergence. 



Physiology of Vision 339 

Why is it that while the orbits diverge the eyes look straight 
ahead? 

By the action of the extra ocular muscles, and especially the 
internal recti. These are the strongest of all the muscles and 
easily overcome the divergence of the orbits. 



When is an eye orthophoric? 

When there is no imbalance of the extra ocular muscles. 



What is meant by acuity of vision and what is the usual stand- 
ard of the same? 

Acuteness of vision is a function of the nervous system 
of the eye ; it is for the retina what tactile sensibility is for the 
skin, ahd the two functions are determined in a similar manner. 
We seek in both for the smallest distance between two points 
which can be perceived separately. For the skin, the mechanical 
pressure of two points of a compass is employed; while for the 
retina it depends upon the retinal images of two luminous points. 

Visual acuity therefore represents the smallest retinal image, 
the form of which can be distinguished, for which its two points 
must be separated by a certain small distance, and this distance 
corresponds in the normal eye to a visual angle of one minute. 



Why does not the amplitude of the convergence fall off as rapidly 
with age as does the amplitude of the accommodation? 

It is probable that all the muscles of the body show some 
loss of power with the advance of age, but the more rapid loss 
of accommodation is due rather to the increasing firmness of 
the crystalline lens than to the loss of power of the ciliary muscle, 
and as a result the former does not respond to the latter. It is 
fair to conclude that the ciliary muscle does not weaken any 
faster than the muscles of convergence, but the first is no longer 
able to increase the convexity of a crystalline which has become 
dense and hard, whereas the second simply causes motion and 
there is no change occurring in the eye to make that motion more 
difficult. 



340 State Board Examinations 

What is meant by the term ''axis of the eye,'' and has the human 
eye an axis? 

The human eye anatomically speaking does not have an 
axis, but the term "axis of the eye" is an imaginary line which 
passes perpendicularly through the center of the cornea, the 
center of the crystalline lens and the center of the fundus, which 
latter is a little to the nasal side of the macula lutea. The front 
end of this line at the apex of the cornea is the anterior pole, and 
the other end at the center of the fundus Is the posterior pole. 



What is abduction of the eyes and when is it exerted and by what 
muscles? 

Abduction is turning of the eyes outwards, and takes place 
when looking out at distant objects after the eyes have been 
converged in near vision and Is effected by the external recti 
muscles. It may be considered a passive function, in contrast 
with adduction which is an active one. 

When prisms are placed before the eyes bases in abduction 
Is called into action to prevent diplopia. 



How can one see opacities in his own eyes if any he present? 

Any object In the eye In front of the sensitive retina inter- 
cepts the light that passes through the pupil and throws shadow^s 
which, under certain conditions, can be perceived. A flame at a 
distance of fifteen feet Is looked at through a strong convex lens 
held two or three Inches from the eye, when a bright patch of light 
will be seen formed by circles of diffusion, upon which the presence 
of any opacities in the eye becomes manifest by the shadows that 
are thrown upon it. 

What is the difference between the optical and visual axis of 
the eye? 

The optic axis is an imaginary line that passes perpendicu- 
larly through the center of the cornea, the center of the crystalline 
lens and the center of the fundus, which point is usually near the 
Inner margin of the macula. 



Physiology of Vision v341 

The visual axis is an imaginary line that passes from the 
object looked at through the nodal point to the macula. 

It is possible that the visual axis should coincide with the 
optic axis, but it seldom does; the two axes crossing and forming 
at the nodal point of the angle of Alpha or of Gamma, 



What advantage has binocular vision over the monocular form, 
as possessed, for instance, hy a one-eyed man? 

The advantages of binocular vision are the power to recog- 
nize depth and perspective or the appreciation of solidity; and 
the more correct estimation of distance as indicated by the amount 
of convergence necessary. 



What is the blind spot and why is it so called? How may its 
location be detected? 

The blind spot is at the entrance of the optic nerve, and it is 
so called because of the absence of the percipient layer of the retina 
there. It is insensitive to light. 

It can be detected by the following experiment: two black 
spots on a white card, left eye closed and with right eye look at 
left hand spot. The card is moved farther from or closer to the eye 
until a position is found where the right hand image is lost, because 
its image falls upon the blind spot. As the position of the card is 
varied, the spot again comiCs into view. 



Why is the associatioyi of accommodation and convergence a 
close one? 

Because it is necessary to use both in equal proportion in near 
vision, and because Nature made it so, both functions being sup- 
plied by the third cranial nerve. 



Why does the pupil look black when it is really transparent? 

Because of the absence of intra-ocular illumination, and 
because there are no rays of light passing from the eye of the 
patient to the eye of the observer. Rays of light entering an eye 



342 State Board Examinations 

are reflected back to their source, and in order to obtain an illu- 
minated pupil these returning rays must be intercepted. 

This is accomplished by making a mirror the source of light, 
and as the rays from the observed eye return to the mirror, the 
observer is able to receive some of these return rays through the 
opening in the mirror. 



What is the result of contraction of the radiating muscular 
fibers of the iris? 

Dilatation of the pupil. 



Pathological Conditions 

What are the different forms of color blindness? 

The different forms of color blindness are red, green and 
violet, of which the first two are the most common. 



How can we tell whether there is any sense of light in a stis- 
pected case of cataract? 

By means of a lighted candle in a darkened room, and by the 
ability of the patient to recognize the light and locate the direction 
from which it comes. 

How can yoii tell if a patient has glaucoma or not? What do you 
see with the ophthalmoscope that will give you this information? 

By the rapid recession of the near point, by the increased 
tension and stony hardness of the eyeball, by the subjective 
symptom of halo or rainbow around a light, by severe neuralgic 
pains that are complained of, by sluggish or immobile pupil, by 
impairment of vision, by cloudiness of the aqueous and vitreous 
and cornea, and by contraction of the field of vision. The ophthal- 
moscope will show venous congestion, arterial pulsation and cup- 
ping of the optic disk. 

What is the difference between an amblyopic eye and an 
asthenopic eye? 

In the amblyopic eye vision is impaired and no improvement 
is afforded by glasses. In the asthenopic eye, vision is usually 
good, but the eyes can be employed but little for close work, 
because all such use of the eye is attended with discomfort or 
pain. An amblyopic eye has blurred vision, while an asthenopic 
eye means weak sight. 

What is meant by retinal fatigue, and how is its existence 
generally proved? 

343 



344 State Board Examinations 

This means exhaustion of the optic nerve and retina, and 
quickly comes on when the eyes are steadily fixed on one object, 
and manifests itself by impairment of vision. 

This is brought home to the optometrist in his daily work. 
The patient is looking intently at the test letters as one lens after 
another is tried ; he is unable to decide which lens is the better, in 
fact he will probably say he cannot see the letters at all. A closing 
of the eye or turning in another direction suffices to relieve the 
exhaustion of the retina and allows the examination to proceed. 
The optometrist would do well to take advantage of these rests 
frequently during his examination. 



What is the effect on the mobility of the eye if the internal rectus 
is paralyzed? 

Partial or total loss of power to turn eye inward, thus 
impairing the function of convergence. 



wse an accident happejts to the eye whereby some of the 
aqueous humor and some of the vitreous humor is lost. Will nature 
replace these humors? 

The aqueous humor will be quickly replaced, but the vitreous 
never. 

In color blindness of the usual type, why does the patient confuse 
shades of green with shades of red? 

According to the Young-Helmholtz theory u'e have three 
primary color perceptions corresponding to the three primary 
colors of red, green and violet. The absence or impairment of 
one or more of the primary perceptions constitutes color blind- 
ness, as a result of which these colors are confused. These 
color sensations are conv-eyed to the brain by three sets of nerves, 
each set conveying not only the sensation of its special color, 
but also to a slight extent that of the other two. 

In color blindness there is a loss of sensation to one or more 
of the three colors, or an impairment of one or more of the sets 
of nerves, which is discovered when the patient is asked to match 
various colored skeins of varus. 



Pathological Conditions 345 

What explanation cafi be given for the oscillation of the eyes in 
many albinos? 

This condition, to which the term nystagmus has been 
given, is most commonly found in persons with congenitally 
defective vision, among whom albinos are to be classed. On 
account of the lack of pigment in the choroid and iris, there is 
no provision for modifying the brightness of the light, and the 
patient suffers from the dazzling, and his vision is impaired. 
Just how the oscillation is produced is a disputed question, but 
it is probable that in these congenital cases the absence of the 
stimulus which distinct retinal images cause, interferes with the 
development and functioning of the centers in the brain that 
govern the co-ordination of the ocular muscles. 

It is an interesting fact that in those patients Avhere the 
nystagmus is due to a congenital defect, there is no complaint 
of oscillation of objects looked at; whereas, when the affection 
comes on later in life, such patients are very much annoyed by 
this symptom. 



What is amblyopia and what are some of the causes of it? 

Amblyopia is impaired vision, which is not caused by an 
error of refraction, but is due to functional disturbance or disease 
of some part of the visual apparatus, either the retina, the optic 
nerve or the brain. It sometimes exists without any evidences 
that are visible to the ophthalmoscope. 

Amblyopia ex anopsia is due to non-use of one eye as in 
strabismus. Reflex amblyopia to irritations in some other part 
of the body. Traumatic amblyopia to injury. Uraemic amblyopia 
to Bright's disease. Toxic amblyopia to tobacco and alcohol in 
excess. 

It may be recognized by its inability to respond to any 
glass placed before the eye, and by the failure of the pin hole to 
afford any improvement in vision. 



Why is it, after the removal of a cataract, that a strong glass 
can be placed over the aphakic eye and be worn with comfort, when 
this cannot be done in cases of high anisometropia? 



346 State Board Examinations 

This is possible only when the other eye suffers with im- 
paired vision, probably from a partly developed cataract. Under 
such conditions the burden of vision is borne by the aphakial eye 
with its proper correction. As long as the vision of one eye is 
good, or at least serviceable, it is not customary to perform a 
cataract operation on the other eye, but a time finally comes when 
vision has been so much impaired that something must be done, 
and then the poorest eye is operated upon. If successful the 
vision of this eye with its correcting lens is so much better than 
the other eye, that it becomes the dominant eye, and is not 
disturbed by the relatively poorer vision of the other eye. 



When is detachment of the retina to he suspected? 

The diagnosis of this condition can be made only by the 
ophthalmoscope. In a partial detachment the normal reflex is 
absent at the portion detached, and instead it is grayish or whitish 
on account of opacity of the retina. As the detached portion 
projects into the cavity of the eyeball, it can be examined by 
the direct method with a convex lens. The gray reflection can 
be seen to be folded, and the few retinal vessels that pass over 
it have lost their light streak, appear dark in color and pursue a 
tortuous course. When the eye is quickly moved in different 
directions, motion can be observed in the folds depending upon 
the amount of the underlying fluid. 



What is the effect of atropine? 

To suspend accommodation and dilate the pupil. It 
paralyzes the peripheral cells of the third cranial nerve as supplied 
to the ciliary muscle and the iris, while it stimulates the radiating 
muscular and sympathetic fibers of the iris. 



What is meant hy night blindness? 

This is a well recognized symptom of the disease known as 
retinitis pigmentosa. While vision is never perfect, yet such a 
patient can see fairly well in bright daylight, but on a dull day 
or at twilight or by insufficient artificial light, vision is very 



Pathological Conditions 347 

greatly impaired. This is said to be due to defective power of 
adaptation of the retina, rather than to defective light sense. 



What is cataract and what is the popular conception of it? 

Cataract is a translucent or opaque crystalline lens. It is 
sometimes spoken of as "on the eye," when it is confused with 
an opacity of the cornea. 



Why may a chalazion temporarily change the refractive con- 
dition of the eye? 

This is a small tumor of the lid and it is possible that by 
pressure on the cornea it may so change its curvature as to pro- 
duce a temporary condition of astigmatism. 



75 there a difference between color blindness and color ignorance? 

Color blindness is a congenital condition, and is due to the 
absence or deficiency of one of the retinal color elements; while, 
in color ignorance there is an inability to name colors because 
of lack of training or education in them. 



The 

Principles of Refraction 

in the Human Eye, Based on the 
Laws of Conjugate Foci 

By SWAN M. BURNETT, M.D., PH.D. 

Formerly Professor of Ophthalmology and Otology in the Georgetown 
University Medical School ; Director of the Eye and Ear Clinic, 
Central Dispensary and Emergency Hospital; Ophthalmo- 
logist to the Children's Hospital and to Providence 
Hospital, etc., Washington, D. C. 

IN this treatise the student is given a condensed but 
thorough grounding in the principles of refraction 
according to a method which is both easy and funda- 
mental. The few laws governing the conjugate foci 
lie at the basis of whatever pertains to the relations of 
the object and its image. 

To bring all the phenomena manifest in the refraction of 
the human eye consecutively under a common explanation by 
these simple laws is, we believe, here imdertaken for the first 
time. The comprehension of much which has hitherto seemed 
difficult to the average student has thus been rendered much 
easier. This is especially true of the theory of Skiascopy, 
which is here elucidated in a manner much more simple and 
direct than by any method hitherto oflrered. 

The authorship is sufficient assurance of the thorough- 
ness of the work. Dr. Burnett was recognized as one of the 
greatest authorities on eye refraction, and this treatise may 
be described as the crystallization of his life-work in this field. 
The text is elucidated by 24 original diagrams, which 
were executed by Chas. F. Prentice, M.E., whose pre-emi- 
nence in mathematical optics is recognized by all ophthalmol- 
ogists. 

BOUND IN SILK CLOTH 
Sent postpaid to any part of the world on receipt of price 

$1.50 

Published by 

THE KEYSTONE PUBLISHING COMPANY 

p. O. BOX 1424 PHILADELPHIA, U. S. A. 



The Optician's Manual 

Volume I. 

By C. H. BROWN, M. D. 

Graduate University of Pennsylvania; Professor of Principles and 

Practice of Optometry; formerly Physician to the Philadelphia 

Hospital; Author of ''Clinics in Optometry," etc. 

THE OPTICIAN'S MANUAL, Vol. I, is the most 
popular and useful work on practical refraction 
ever written, and has been the entire optical edu- 
cation of many hundred successful refractionists. 
The knowledge it contains was more effective in building up 
the profession of optometry than any other educational 
factor. It is, in fact, the foundation structure of all op- 
tometric knowledge as the titles of its ten chapters show : 

Chapter I. —Introductory Remarks. 
Chapter II.— The Eye Anatomically. 
Chapter HI.— The Eye Optically; or, 

The Physiology of Vision. 
Chapter IV.— Optics. 
Chapter V.— Lenses. 
Chapter VI. — Numbering of Lenses. 
Chapter VIL— The Use and Value of Glasses. 
Chapter VIII.— Outfit Required. 
Chapter IX. — Method of Examination. 
Chapter X. — Presbyopia. 

In its present revised and enlarged form this volume is 
the recognized standard text-book on practical refraction, 
being used as such in all schools of Optics. A study of it 
is essential to an intelligent appreciation of its companion 
treatise. The Optician's Manual, Vol. 11. A comprehensive 
index adds much to its usefulness to both student and 
practitioner. 

Bound in cloth— 422 pagfes— colored plates and illustrations. 
Sent postpaid to any part of the world on receipt of price 

$2.50 

Published by 

THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



The Optician's Manual 

VOL. II. 

By C. H. BROWN, M.D. 

Graduate University of Pennsylvania ; Professor of Optics and Refraction ; 

Formerly Physician in Philadelphia Hospital ; Author of "Clinics in 

Optometry"; "State Board Questions and Answers "; Etc. 

THE Optician's Manual, Vol. II., is a direct 
continuation of The Optician s Manual, Vol. I., 
being a much more advanced and compre- 
hensive treatise. It covers in minutest detail 
the four great subdivisions of practical eye refraction, viz : 

MYOPIA 

HYPERMETROPIA 
ASTIGMATISM 
MUSCULAR ANOMALIES 

It contains the most authoritative and complete 
researches up to date on these subjects, treated by the 
master hand of an eminent oculist and optical teacher. 
It is thoroughly practical, explicit in statement and 
accurate as to fact. All refractive errors and complica- 
tions are clearly explained, and the methods of correc- 
tion thoroughly elucidated. 

This book fills the last great want in higher refrac- 
tive optics, and the knowledge contained in it marks 
the standard of professionalism. 

Bound in Cloth. 408 pages, with illustrations. 

Sent postpaid to any part of the world on receipt of price 

$2.50 



Published by 
THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



Clinics in Optometry 

By C. H. BROWN, M. D. 

Graduate University of Pennsylvania : Professor of Principles and 

Practice of Optometry ; formerly Physician to the 

Philadelphia Hospital ; Author of the 

Optician's Manual, Etc. 



CLINICS IN OPTOMETRY" is a unique 
work in the field of practical refraction and 
fills a want that has been seriously felt both 
by oculists and optometrists. 
The book is a compilation of optometric cUnics, each 
cHnic being complete in itself Together they cover 
all manner of refractive eye defects, from the simplest 
to the most complicated, giving in minutest detail the 
proper procedure to follow in the diagnosis, treatment 
and correction of all such defects. 

Practically every case that can come before you 
is thoroughly explained in all its phases in this useful 
volume, making mistakes or oversights impossible and 
assuring correct and successful treatment. 

The author's experience in teaching the science of 
refraction to thousands of pupils peculiarly equipped 
him for compiling these clinics, all of which are actual 
cases of refractive error that came before him in his 
practice as an oculist. 

A copious index makes reference to any particular 
case, test or method, the work of a moment. 

BOUND IN SILK CLOTH 
Sent postpaid to any part of the ^vorld on receipt of price 

$2.50 

A* 



Published by 
THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



Tests and Studies 

of the Ocular 

Muscles 

By ERNEST E. MADDOX, M.D., F.R.C.S., Ed. 

Ophthalmic Surgeon to the Royal Victoria Hospital, Bournemouth, England ; 
formerly Syme Surgical Fellow, Edinburgh University 

THIS book is universally recognized as the 
standard treatise on the muscles of the eye, 
their functions, anomalies, insufficiencies, tests 
and optical treatment. 
All optometrists recognize that the most troublesome 
subdivision of refractive work is muscular anomalies. 
Even those who have mastered all the other intricacies 
of visual correction will often find their skill frustrated 
and their effi3rts nullified if they have not thoroughly 
mastered the ocular muscles. 

The eye specialist can thoroughly equip himself 
in this fundamental essential by studying the work of 
Dr. Maddox, who is known in the world of medicine 
as the greatest investigator and authority on the sub- 
ject of eye muscles. 

The present volume is the second edition ot the 
work, specially revised and enlarged by the author. It 
is copiously illustrated and the comprehensive index 
greatly facilitates reference. 

Bound in Silk Cloth — 261 Pages — 110 Illustrations. 
Sent postpaid to any part of the world on receipt of price 

$2.50 

Published by 

THE KEYSTONE PUBLISHING COMPANY 

p. O. BOX 1424 PHILADELPHIA, U. S. A. 



Physiologic Optics 

Ocular Dioptrics — Functions of the 

Retina — Ocular Movements and 

Binocular Vision 

By DR. M. TSCHERNING 

Director of the Laboratory of Ophthalmology 
at the Sorbonne, Paris 

AUTHORIZED TRANSLATION 

By CARL WEILAND, M.D. 

Former Chief of Clinic in the Eye Department of the 
Jefferson College Hospital, Philadelphia, Pa. 

THIS book is recognized in the scientific and medical 
world as the one complete and authoritative treatise 
on physiologic optics. Its distinguished author is 
admittedly the greatest authority on this subject, 
and his book embodies not only his own researches, but those 
of the several hundred investigators who, in the past hundred 
years, made the eye their specialty and life study. 

Tscherning has sifted the gold of all optical research 
from the dross, and his book, as now published in English, 
with many additions, is the most valuable mine of reliable 
optical knowledge within reach of ophthalmologists. It con- 
tains 380 pages and 212 illustrations, and its reference list 
comprises the entire galaxy of scientists who have made 
the century famous in the world of optics. 

The chapters on Ophthalmometry, Ophthalmoscopy, 
Accommodation, Astigmatism, Aberration and Entoptic 
Phenomena, etc. — in fact, the entire book contains so much 
that is practical and necessary, that no refractionist can 
afford to be without it. 

Bound in Cloth. 380 Pages, 212 Illustrations. 

Sent postpaid to any part of the world on receipt of price 

$3.00 



Published by 

THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



The Refractive and 

Motor Mechanism 

of the Eye 

By WILLIAM NORWOOD SOUTER, M.D. 

Associate Ophthalmologist, Episcopal Eye, Ear 
and Throat Hospital, Washington, D. C. 

THIS work by one of the most eminent ophthal- 
mologists in the United States, brings the 
science of eye refraction right up to date and 
embodies, in addition to the profound knowledge of 
the author, all the researches on the subject that experi- 
ence has established as authoritative. 

The geometric and mathematical optics on which 
the principles of optometry are based, necessary infor- 
mation to optical students of to-day, will be found in 
simplified form in the Appendix of this treatise. 

Students, teachers and practitioners alike, in study- 
ing this book or using it for reference, have the assur- 
ance of absolute reliability of statement and complete 
elimination of the misleading fallacies which mar the 
worth of many works on this subject. 

It contains 350 pages with 148 illustrations, many 
entirely original, and is probably the only scientific 
work ever published in which every single reference 
was verified absolutely by the author himself. 

Sent postpaid to any part of the world on receipt of price 

$2.50 

Published by 
THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



ophthalmic Lenses 

Dioptric Formulae for Combined Cylindrical 

Lenses, The Prism-Dioptry and 

Other Original Papers 

By CHARLES F. PRENTICE, M.E. 

A new and revised edition of all the original papers of this noted author, 
combined in one volume. In this revised form, with the addition of recent 
research, these standard papers are of increased value. Combined in one 
volume, they are the greatest compilation on the subject of lenses extant. 
This book of over 200 pages contains the following papers: 

Ophthalmis Lenses. 

Dioptric Formulae for Combined Cylindrical Lenses. 

The Prism-Dioptry. 

A Metric System of Numbering and Measuring Prisms. 

The Eelaiiou of the Prism-Dioptry to the .Meter Angle. 

The Relation of the Prism-Dioptry to the Lens-Dioplry. 
The Perfected Prismometer. 
The Prismometric Scale. 
On the Practical Execution of Ophthalmic Prescriptions 

involving Prisms. 
A Problem in Cemented Bi-Focal Lenses, Solved by the 

Prism-Dioptry. 
Why Strong Contra-Generic Lenses of Equal Power Fail 

to Neutralize Each Other. 
The Advantages of the Sphero-Toric Lens. 
The Iris, as Diaphragm and Photostat. 
The Typoscope. 
The Correction of Depleted Dynamic Refraction (Presbyopia). 

PRESS NOTICES OF THE ORIGINAL EDITION: 

OPHTHALMIC LENSES 

"The work stands alone, in its present form, a compendium of the various laws of 
physics relative to this subject that are so difficult of access in scattered treatises." 

—yen- E)igland Medical Gazette. 

" It is the most complete and best illustrated book on this special subject ever published." 

^Hoiological Review, New York. 

"Of all the simple treatises on the properties of lenses that we have seen, this is incom- 
parably the best. . . . The teacher of the average medical student will hail this 
little work as a great boon." — Archives of Ophthalmology, edited by H. Knapp, M.D. 

Bound in Silk Cloth. 110 Original Diagrams. 

Sent postpaid to any part of the world on receipt of price 

$2.00 

Published by 

THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U- S. A. 



The Making of a 

Mechanical 

Optician 

A PRACTICAL TREATISE ON THE MECHANICAL 
WORK OF OPTOMETRISTS AND OPTICIANS 

By W. W. SLADE 

Of the Globe Optical Company, Boston, Mass. 

THIS new volume is the only treatise published 
on mechanical optics with special reference to 
the work of optometrists and opticians. The 
author is regarded in the optical world as the 
highest authority on this important branch of optical prac- 
tice, and his reputation is an assurance to the trade of the 
absolute reliability of the work. 

The seventeen chapters of the new work cover thor- 
oughly all the mechanical operations of the optician from in- 
forming him as to the machinery needed and the handling 
of tools, to the final operation in the correct filling of 
prescriptions and the production of first-class optical work. 
There are chapters on lens marking and cutting, lens grind- 
ing, drilling and mounting, bifocal work, soldering, repair- 
ing, bridge bending, surface grinding, etc., etc. 

The illustrations are in greater part entirely original 
and were especially executed for the purposes of this 
volume. These add greatly to the instructive character of 
the book, which is indispensable to every optometrist and 
optician. 

Send postpaid to any part of the world on receipt of price 

$2.50 



Published by 

THE KEYSTONE PUBLISHING COMPANY 

p. O. BOX 1424 PHILADELPHIA, U. S. A. 



Record-Book of 
Optometric 
Examinations 

A RECORD-BOOK, wherein to record opto- 
/ ^ metric examinations, is an indispensable ad- 
/ ^ junct to an optometrist's outfit. 

The Keystone Record-Book of Optometric Exami- 
nations was specially prepared for this purpose. It 
excels all others in being not only a record-book, but 
an invaluable guide in examination. 

The book contains two hundred record forms with 
printed headings, suggesting, in the proper order, the 
course of examination that should be pursued to obtain 
most accurate results. 

Each book has an index, which enables the optom- 
etrist to refer instantly to the case of any particular 
patient. 

The Keystone Record-Book diminishes the time 
and labor required for examinations, obviates possible 
oversights from carelessness, and assures a systematic 
and thorough examination of the eye, as well as 
furnishing a permanent record of all examinations. 

Sent postpaid to any part of the world on receipt of price 

$2.00 



Published by 
THE KEYSTONE PUBLISHING COMPANY 

P. O. BOX 1424 PHILADELPHIA, U. S. A. 



